Interesting Probability Problems (p Nitin)

Probability of an event: If there are n elementary events associated with a random experiments and m of them are favourableto an event A, then the probability of happening or occurence of A denoted by P(A) and is defined as the ratio m/n

Thus, P(A) = m/n

Let us use this formula and solve some interesting probability problems.

Pro 1:Two dice are thrown simultaneously. Find the probability of getting :

a)an even number as the sum

b)the sum of prime number

c)a total of at least 10

d)a doublet of even number

e) a multiple of 2 on one dice and a multiple of 3 on the other

f)same number on both dice

g)a multiple of 3 as the sum

Solution: When two dice are thrown together the sample space S associated with the random experiment is given by

S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Clearly total number of events is 36

a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,

A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}

Favourable number of elementary events = 18

So, required probability = 18/36 =1/2

b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,

A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}

Favourable number of elementary events = 15

So, required probability = 15/36 = 5/12

c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,

A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36=1/6

d)Let A be the event of getting a double of an even number.Then ,

A={(2,2),(4,4),(6,6)}

Favourable number of elementary events= 3

So, required probability=3/36=1/12

e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,

A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}

Favourable number of elementary events= 11

So, required probability=

f) Let A be the event of "getting the same number on both the dice."Then,

A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36 = 1/6

g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}

Favourable number of elementary events= 12

So, required probability=12/36 = 1/3

Some interesting probability problems:

Pro 2: Find the probability that a leap year will contain 53 sundays.

Sol: In a leap year there are 366 days.

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays.The remaining 2 days can be:

(i) Sunday and Monday, (ii) Monday and Tuesday,

(iii)Tuesday and Wednesday , (iv) Wednesday and Thursday,

(v)Thursday and Friday,(vi)Friday and Saturday,

(vii)Saturday and Sunday.

If S is the sample space associated with this problem , then S consists of the above seven points.

The total number of elementary events =7

Let A be the event that a leap year has 53 Sundays.In order that a leap year, selected at random, should have 53 Sundays, one of the "over" days must be a Sunday.This is can be in any one of the following two ways

(i) Sunday and Monday or (ii)Saturday and Sunday

Favourable number of elementary events =2

Hence, required probability = 2/7

Some more interesting probability problems:

Pro 3: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase.

Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens

Therefore, required probability = 1/5040

Processing ...

Interest Rate Differential (p Nitin)

A differential measuring the gap in interest rates between two similar interest-bearing assets is called as interest rate differential.

The term interest rate differential (IRD) has multiple usages.

In the mortgage industry, an Interest Rate Differential is used to determine a pre-payment penalty, to account for the excess of the paid-off mortgage's rate over prevailing rates.

In foreign exchange, the term refers to the practice of borrowing in the currency of a country where interest rates are low and lending the proceeds in the currency of a country where interest rates are higher. The difference in borrowing and lending rates, i.e. the Interest Rate Differential, is anticipated profit.

Applications of Interest Rate Differential in mortgage industry

Prepayment

Pre-payment of a mortgage in the first year usually results in a fee proportionate to the size of the mortgage. Interest rates do not generally change much in one year. The amount lost in future interest collections from a one-year drop in interest rates does not usually approach the usual balance-based first-year pre-payment penalties. Therefore it is reasonable for lenders not to use the IRD as a base for such fees.

After the first year of a mortgage, a pre-payment penalty may be based on the change in interest rates since the loan origination. The remainder of the fixed-rate term (before the loan resets in its ARM phase) would be the basis for determining the bank's loss. For example, if an ARM resets after five years, and the mortgage is repaid in full after three, the lender would charge for the interest income lost. If the fixed rate was 8 percent and prevailing rates are currently 6 percent, then the lender would charge for the 2 percent difference to be lost over the next two years.

A homeowner that thinks he is taking advantage of prevailing rates in fact may not be, at least for the remainder of the mortgage's fixed-rate term.

Interest modification

Instead of paying the loan off in full, a homeowner in an ARM may want the interest adjusted before the end of the term. This may be a futile exercise if the bank charges for the interest income to be lost over the subsequent two years. The situation is the same as in a mortgage payoff, trying to recover the drop in interest rates with an up-front fee.

Application of Interest Rate Differential in foriegn excahnge

Forex: Carry Trade Market

The carry trade market, in foreign exchange, is the practice of borrowing in the currency of a country where interest rates are low and lending the proceeds in the currency of a country where interest rates are higher. The Interest Rate Differential determines the profit. This profit is not guaranteed, however, as exchange rates do fluctuate. The practice is also known as uncovered interest arbitrage.

Traders in the foreign exchange market use interest rate differentials (IRD) when pricing forward exchange rates. Based on the interest rate parity, a trader can create an expectation of the future exchange rate between two currencies and set the premium (or discount) on the current market exchange rate futures contracts.

The Interest Rate Differential is a key component of the carry trade. For example, say an investor borrows US$1,000 and converts the funds into British pounds, allowing the investor to purchase a British bond. If the purchased bond yields 7% while the equivalent U.S. bond yields 3%, then the IRD equals 4% (7-3%). The Interest Rate Differential is the amount the investor can expect to profit using a carry trade. This profit is ensured only if the exchange rate between dollars and pounds remains constant.

Forex: Futures

In interest arbitrage, an exchange-rate fluctuation can effectively cancel out profit from the Interest Rate Differential. Therefore, a forward contract (or future) may be sold to fix the exchange rate for the future repatriation of the second currency to the first. This is called hedged interest arbitrage, and is virtually risk-free, unless a government institutes strong controls that prevent fulfillment of the futures contract or repatriation of the investment returns.

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Interesting Probability Problems (p Nitin)

Probability of an event: If there are n elementary events associated with a random experiments and m of them are favourableto an event A, then the probability of happening or occurence of A denoted by P(A) and is defined as the ratio m/n

Thus, P(A) = m/n

Let us use this formula and solve some interesting probability problems.

Pro 1:Two dice are thrown simultaneously. Find the probability of getting :

a)an even number as the sum

b)the sum of prime number

c)a total of at least 10

d)a doublet of even number

e) a multiple of 2 on one dice and a multiple of 3 on the other

f)same number on both dice

g)a multiple of 3 as the sum

Solution: When two dice are thrown together the sample space S associated with the random experiment is given by

S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Clearly total number of events is 36

a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,

A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}

Favourable number of elementary events = 18

So, required probability = 18/36 =1/2

b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,

A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}

Favourable number of elementary events = 15

So, required probability = 15/36 = 5/12

c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,

A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36=1/6

d)Let A be the event of getting a double of an even number.Then ,

A={(2,2),(4,4),(6,6)}

Favourable number of elementary events= 3

So, required probability=3/36=1/12

e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,

A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}

Favourable number of elementary events= 11

So, required probability=

f) Let A be the event of "getting the same number on both the dice."Then,

A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36 = 1/6

g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}

Favourable number of elementary events= 12

So, required probability=12/36 = 1/3

Some interesting probability problems:

Pro 2: Find the probability that a leap year will contain 53 sundays.

Sol: In a leap year there are 366 days.

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays.The remaining 2 days can be:

(i) Sunday and Monday, (ii) Monday and Tuesday,

(iii)Tuesday and Wednesday , (iv) Wednesday and Thursday,

(v)Thursday and Friday,(vi)Friday and Saturday,

(vii)Saturday and Sunday.

If S is the sample space associated with this problem , then S consists of the above seven points.

The total number of elementary events =7

Let A be the event that a leap year has 53 Sundays.In order that a leap year, selected at random, should have 53 Sundays, one of the "over" days must be a Sunday.This is can be in any one of the following two ways

(i) Sunday and Monday or (ii)Saturday and Sunday

Favourable number of elementary events =2

Hence, required probability = 2/7

Some more interesting probability problems:

Pro 3: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase.

Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens

Therefore, required probability = 1/5040

Processing ...

Interest Rate Differential (p Nitin)

A differential measuring the gap in interest rates between two similar interest-bearing assets is called as interest rate differential.

The term interest rate differential (IRD) has multiple usages.

In the mortgage industry, an Interest Rate Differential is used to determine a pre-payment penalty, to account for the excess of the paid-off mortgage's rate over prevailing rates.

In foreign exchange, the term refers to the practice of borrowing in the currency of a country where interest rates are low and lending the proceeds in the currency of a country where interest rates are higher. The difference in borrowing and lending rates, i.e. the Interest Rate Differential, is anticipated profit.

Applications of Interest Rate Differential in mortgage industry

Prepayment

Pre-payment of a mortgage in the first year usually results in a fee proportionate to the size of the mortgage. Interest rates do not generally change much in one year. The amount lost in future interest collections from a one-year drop in interest rates does not usually approach the usual balance-based first-year pre-payment penalties. Therefore it is reasonable for lenders not to use the IRD as a base for such fees.

After the first year of a mortgage, a pre-payment penalty may be based on the change in interest rates since the loan origination. The remainder of the fixed-rate term (before the loan resets in its ARM phase) would be the basis for determining the bank's loss. For example, if an ARM resets after five years, and the mortgage is repaid in full after three, the lender would charge for the interest income lost. If the fixed rate was 8 percent and prevailing rates are currently 6 percent, then the lender would charge for the 2 percent difference to be lost over the next two years.

A homeowner that thinks he is taking advantage of prevailing rates in fact may not be, at least for the remainder of the mortgage's fixed-rate term.

Interest modification

Instead of paying the loan off in full, a homeowner in an ARM may want the interest adjusted before the end of the term. This may be a futile exercise if the bank charges for the interest income to be lost over the subsequent two years. The situation is the same as in a mortgage payoff, trying to recover the drop in interest rates with an up-front fee.

Application of Interest Rate Differential in foriegn excahnge

Forex: Carry Trade Market

The carry trade market, in foreign exchange, is the practice of borrowing in the currency of a country where interest rates are low and lending the proceeds in the currency of a country where interest rates are higher. The Interest Rate Differential determines the profit. This profit is not guaranteed, however, as exchange rates do fluctuate. The practice is also known as uncovered interest arbitrage.

Traders in the foreign exchange market use interest rate differentials (IRD) when pricing forward exchange rates. Based on the interest rate parity, a trader can create an expectation of the future exchange rate between two currencies and set the premium (or discount) on the current market exchange rate futures contracts.

The Interest Rate Differential is a key component of the carry trade. For example, say an investor borrows US$1,000 and converts the funds into British pounds, allowing the investor to purchase a British bond. If the purchased bond yields 7% while the equivalent U.S. bond yields 3%, then the IRD equals 4% (7-3%). The Interest Rate Differential is the amount the investor can expect to profit using a carry trade. This profit is ensured only if the exchange rate between dollars and pounds remains constant.

Forex: Futures

In interest arbitrage, an exchange-rate fluctuation can effectively cancel out profit from the Interest Rate Differential. Therefore, a forward contract (or future) may be sold to fix the exchange rate for the future repatriation of the second currency to the first. This is called hedged interest arbitrage, and is virtually risk-free, unless a government institutes strong controls that prevent fulfillment of the futures contract or repatriation of the investment returns.

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Joint Probability (p Nitin)

Probability in math is defined as the chance of happening something in future. The two random variables A and B are defined on the same probability space, the joint probability distribution for A and B defines the probability of events defined in terms of both A and B. In the case of having only two random variables, this is called a bi-variate distribution, but the thought simplify to any number of random variables, giving a multivariate distribution.

Joint probability:

A numerical measure where the chance of two events happening together and at the same time are calculated. For example if the probability of event B happening at the same time event A happens, then the Joint probability has been given as follows.
Joint probability notation takes the form:

P (A 'nn' B) or P (A and B)

Indicates the joint probability of A and B.

Example: The joint probability can be calculated by rolling a 2 and a 5 with two dissimilar dice.

with and without replacemant-Joint probability:

Joint probability is used in multistage testing .Joint probability can be done with replacement or without replacement.

With replacement: It indicates that the thing that are chosen on one stage are returned to the sample space before the next choice is made .For example, tossing a head on the first toss does not affect the outcome of flipping the coin a second time.

The probability that independent events A and B occur at the same time can be found by using the multiplication rule, or the product of the entity probabilities.

Example 1:

If you pick two cards from the deck without replacement, find the probability that they will both be aces.

Solution:

Total number of aces in the deck of cards = 4.

Cards picked up = 2 aces.

total number of aces* (total number of aces-1)
Hence the probability = --------------------------------------------------
total number of cards*( total number of cards-1)

P (AA) = '(4/52)*(3/51)' = '1/221' .

Without replacement:

Uses the same idea, if the first choice is not replaced only we consider the change in the sample space. Still we use the multiplication rule, but for each of the stages the numerator and/or denominator decreases

Example 2:

Find the probability of tossing a fair coin twice in a row, getting heads both times.

Solution:

While tossing a fair coin once we get head or tail.

Given that while tossing a coin head occurs,So

We know that the probability =(Number of favourable outcomes/Total number of outcomes)

Therefore probability of getting head while tossing the coin once P(H)= '1/2.'

Similarly tossing a coin next time we have probability of getting head P(H) ='1/2.'

As the question is to find the probability of tossing a fair coin twice in a row, getting heads both times it indicates that we have to find the joint probability without replacement.

As the probability of tossing a head is ' 1/2 ' each time P (H,H) =' (1/2) *(1/2) = 1/4.'

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Accurate Temperature Measurements (p Nitin)

The temperature above or below which pair of partially miscible liquids becomes miscible in all proportions, is called critical solution temperature (CST) or the consulate temperature or accurate temperature for the pair.

Accurate temperature measurements for some Liquids:

Some liquids pair attains complete miscibility above a certain temperature in which case they are said to have the upper critical solution temperature UCST while some liquid pairs show complete miscibility below a certain temperature when they are said to have lower critical temperature LCST. On the other hand, some liquid pairs show both LCST and UCST. All are incompletely miscible liquids pairs have, in general an upper or a LCST. However, the upper CST cannot be realized if one or both the components boil out before it is reached and the lower CST may not be observed if one or both the components freeze out before it is reached. Thus, under atmospheric conditions none of the CST is observed for ether and water system.

Accurate temperature measurements for Atmosphere

The atmosphere is really a composite in its behavior. Just because of this, ruling out the correct elucidation for the behavior of an atmosphere is important as well. Almost all the scientists would agree that a repetition of carbon dioxide in the Earth's atmosphere must have some effect on the temperature of the Earth. But it is certainly less either we all recognize the effect of this increase.

Thus, there are several reasons:

First, thing is that under the influence of a man the doubling amount of carbon dioxide in the atmosphere is little compared to the Earth's natural cooling speed,but this will be only 1%.

Secondly, the most important greenhouse effect. Instead of gas it occupies the water vapor. Water vapor present over the Earth is really changeable, both in space and in time.

Third, ways in which both clouds and water vapor are feedback and at last influence the effect of temperature on the Earth are, at top, badly understood.

Fourth, it includes the entire Earth indeed in a state that scientists describe as "radiative equilibrium," where the external sunlight equals the retiring infrared radiation to provide a approximately steady temperature, thus the surface would be far away from this radioactive balance condition.

Thus, the rate of evaporation and convection processes in the atmosphere transfer heat from the surface to the upper troposphere, where it can be much more capably radiated into space because it is above several of the greenhouse-trapping water vapor. So in squat, it is convective overturn of the atmosphere - poorly represent the computer models of global warming - that primarily determines the temperature distribution of the surface and upper troposphere, not radiation balance.

Factors Affecting the Accurate measurement for Temperature

Factors Affecting the Accurate measurement for Temperature as follows

Nature of the gas
Nature of the solvent.
Temperature of the gas-liquid system.
Pressure of the gas.
It has been found that under a constant pressure, the solubility of a gas diminishes with rise in temperature. The absorption coefficients of a few gases at different temperatures are given below:

CO2 = 0

O2= 0.0484

N2=0.0239 etc.

Thus the mass of a gas dissolved per unit volume of a solvent is proportional to the pressure of the gas in equilibrium with the solution at accurate temperature.

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Some Important Compounds of Calcium (p Nitin)

The important compounds of calcium are

Gypsum and Plaster of paris

Mortar

Gypsum and Plaster of paris :

In nature , calcium sulphate occurs as gypsum and also as anhydrite . Alabaster is finely divided naturally calcium sulphate . The presence of calcium causes permanent hardness to water .

In the laboratory, calcium sulphate is prepared by the action of dil H2SO4 on lime (CaO) or CaCO3 . On concentration of the solution , gypsum CaSO4.2H2O separates out .

CaO + H2SO4 + H2O ---- CaSO4.2H2O

CaCO3 + H2SO4 + H2O ---- CaSO4.2H2O + CO2

Calcium sulphate or gypsum is sparingly soluble in water . Its solubility decreases on heating . It dissolves in solutions like that of ammonium sulphate due to double salt formation . When heated to 120oC to 130oC , a half hydrate or a hemihydrate ( CaSO4. H2O ) is formed . The hemihydrate is often referred to as plaster of paris . Both gypsum and plaster of paris lose water of hydration at 200oC and are converted into anhydrous calcium sulphate , known as "Dead burnt" CaSO4 . This is so called because it does not set with water .

Plaster of paris , when mixed with water ( half of its amount by weight ) becomes a paste which sets to a hard mass on standing . The process is known as "setting" . As the mass becomes hard , its volume increases and finally changes to CaSO4.2H2O .

There are two stages in the setting of plaster of paris . In the first stage , known as "setting stage" , the hemihydrate changes into orthorhombicdihydrite . In the second phase orthorhombic dihydrate is converted into mono-clinic dihydrate , gypsum . This stage is known as "hardening stage" .

Plaster of paris is used in the manufacture of crucibls , models etc . Surgeons use it for setting the fractured bones in position . Dentists also use it in hospitals . Toys are made with it . In the preparation of chalks used in class room teaching , plaster is used .

Gypsum is used in the preparation of cement and plaster of paris .

Mortar is an important compounds of calcium

An intimate mixture of 1 part of slaked lime , 3 parts of sand and water is known as lime "mortar" . Sand makes mortar porous and also harder . While hardening , sand prevents the cracks on contraction . Even though mortar was used in the construction of houses for several decades , it is not clearly known how it becomes so hard . Evaporation of water may be one of the reasons . Slaked lime , when mixed with sand , hardens slowly , due to formation of calcium silicate .

Ca(OH)2 + SiO2 gives CaSiO3 + H2O

Similarly , it may combine with CO2 to form calcium carbonate .

Ca(OH)2 + CO2 gives CaCO3 + H2O

Mortar becomes hard with time due to several chemical reaction taking place at the centre of the mass .

Mortar , mixed with cement is called "cement mortar" . This is stronger than mortar .

Lime stone and clay when heated together give "hydraulic mortar" .This can set like cement on addition of water . This is used for bleaching purposes and as an antiseptic .

Uses of Calcium

It is used as a reducing agent in the extraction of metals like uranium , zirconium and thorium .

It is also used as an alloying agent in the production of beryllium , aluminium , copper , lead and magnesium alloys .

It is also used as an deoxidizers , desulfurizer for various ferrous and non-ferrous alloys .

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Probability Survey (p Nitin)

The probability survey is the way of expressing an event that will occur. The probability survey is the event, the experiments that are repeatedly done under some predefined conditions. The results for one or more experiments are not equal. These types of experiments are called as the random experiments or simply experiments. The probability includes the sample space, trail and different forms of events.

Terms present in the probability survey:

Sample space indicates the total number of possibilities for an experiment.
Trial corresponds to the experiment is performed.
Event specifies the outcome of the experiments.
Exhaustive events are an event which contains all the necessary possible outcomes of the experiment.
Mutually exclusive events are the two events that cannot occur simultaneously.
The probability certain likely defines the equally likely event in the probability. Equally likely event means that the two or more events have an equal probability. For example while tossing the die the probability for getting the tail and also the probability for getting the head are the equally likely events. The equally likely event determines the equal probability for the events.

Example problems for probability survey:

Ex 1 :A jar has 6 gray and 9 red marbles. What is the probability to get one gray marbles from the urn without replacement?

Sol:

The number of marbles in the jar is 6 gray and 9 red marbles.

The total numbers of marbles are 15 marbles.

The possibility for getting a gray ball is 6.

The required probability is 6/15 .

Ex 2 : While tossing a fair die, find the complementary probability of the numbers greater than 3.

Sol:

The sample space for the die is S= {1, 2, 3, 4, 5, 6}

The total number of sample space =6.

A is the event for getting the number greater than 3.

A= {4, 5, 6}

The number of events greater than 3 is n (A) =3

P (A) =n (A)/ n(S)

P (A) = 3/6

P (A) = 1/2

The probability for getting the numbers greater than 3 is 1/2 .

The formula for the complementary probability is 1- P (original probability).

The required probability = 1-P (A)

The required probability = 1- 1/2

The required probability = 1/2

The complementary probability for the numbers greater than 3 is 1/2 .

Survey of probability of certain likely events:

Some examples for probability certain likely:

Probability for getting the head and the tail when a coin is tossed only one time.
The probability for getting the number 3 and number 4 are equally likely events.
If an urn contains 5 white balls and 5 red balls. In that the probability for getting the single white ball and also the probability for getting the single red ball are the equally likely events.

Ex 3 : A jar has 5 gray and 7 green marbles. What is the probability to get one gray marbles and also probability for getting 1 green marbles? Determine whether the above events are equally likely events.

Sol:

The number of marbles in the jar is 5 gray and 7 green marbles.

The total numbers of marbles are 12 marbles.

The possibility for getting a gray marble is 5.

The probability for getting one gray marble is 5/12.

The possibility for getting a green marble is 7.

The probability for getting one green marble is 7/12.

The probabilities are 5/12 and also 7/12. These two probabilities are not the equally likely event because the probability of that two events are not same they are different.

Ex 4 : A single six face die is rolled. Find the probability for getting the number 6 and also 3. Determine whether these two events are equally likely events are not.

Sol:

The sample space for the die is S= {1, 2, 3, 4, 5, 6}

The total number of sample space is 6.

The probability for getting the number 3 is 1/6 .

The probability for getting the number 6 is 1/6 .

The probabilities for the two events are 1/6 and 1/6 respectively. The probabilities for the two events are equal. So these two events are equally likely events.

Practice problems:

Two coins are tossed at the same time. What is the probability to get two tails?
Ans: 1/2 .

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Chemicals in Food (p Nitin)

The food contamination by hazardous chemical is a universal health concern and is a most important cause of every problem globally. The occurrence of contamination is through several environmental pollution or through the intended use of different chemicals, such as pesticides, animal drugs and other agrochemicals.

While in 1976, the WHO implement the Global Environment Monitoring System - Food Contamination Monitoring and Assessment Programme (GEMS/Food), which has educated governments, the Codex Alimentarius Commission and other related institutions, as well as the public, on levels and trend of contaminants in food, their part to total individual exposure, and implication with regard to public health and trade.

Food additives and contaminants resultant from food mechanized and dispensation can also harmfully affect health.

Chemicals in food

A wide range of activities have been developed to estimate the security of food components. The International Programme on Chemical Safety (IPCS) has developed this type of safety measures.

These actions provide the secretariats and technical advice to the Joint authority on Food Additives (JECFA) and the Joint Meeting on Pesticide Residues (JMPR) and carried out worldwide risk assessment of chemical anxiety such as acryl amide, formed as a derivative of food processing and cooking.

For the estimation of chemicals in food, as with other chemicals evaluation work, the development, organization and use of globally established, systematically sound and translucent principles and method is essentially important.

List of Chemicals in foods

Food Chemicals: Acetic acid, citric acid, tartaric acid, meleic acid, gems powder, sorbitol, propyl glycol, calcium propionate, bakery chemicals, flavoring chemicals, potassium meta bisulphate, menthol, peppermint, thymol ammonium bicarbonate, ammonium carbonate, pectin, sodium benzoate, butyrate, hydroxyl, anisole, etc.

General Chemicals in foods: Caustic soda flakes, caustic potash, sodium nitrate, sodium nitrite, sodium sulphate, sodium bicarbonate, soda ash, calcium carbonate, calcium sulphate, zinc sulphate, zinc oxide, hydrochloric acid and hydrofluoric acid

Industrial Chemicals: Caustic soda flakes, potash flakes, nitrate, sodium nitrite, sodium sulphate, sodium bicarbonate, soda ash, calcium carbonate, calcium sulphate, zinc sulphate, zinc oxide, hydrochloric acid, hydrofluoric acid, toluene, phenol, benzene, cyclohexane, carbon tetra chloride, methyl dichloride.

Chemicals used in Paints and Inks: Rosin, turpentine, pine oil, mineral turpentine oil, ethyl silicate, phenol, butyl acetate, ethyl acetate, methanol, propyl glycol and titanium oxide

Chemicals used in Pigment Color: Red, green, beta blue, yellow and pink

Chemicals used as Disinfectant: Instrument cleaner, chlorohexadine, chloramines, benzalkonium chloride, Lysol, sodium benzoate, potassium permanganate, alum, hydrogen peroxide, mouth cleaning chemicals, ethanol and ISO propyl alcohol

Solvent chemicals: Toluene, benzene, phenol, hexane, heptanes, industrial solvent base oil, rubber process oil, M.E.K., M.I.B.K.

Dental Chemicals: The dental chemicals as follows compound of Fluorine, mercury, meleic acid, benzalkonium chloride tartaric acid and calcium sulphate

Leather Chemicals: The leather chemicals as follows sodium nitrate, softener, resorcinol, sodium benzoate and basic color

Water Treatment Chemicals: Alum, poly aluminum chloride, polyelectrolite, sodium hexameta phosphate, disodium phosphate, potassium permanganate, lime powder, sodium hypochlorite, bleaching.

Understand more on about the Definition Nomenclature, and its Illustrations. Between, if you have issue on these Expanded Octet Rule keep verifying my content i will try to help you. Please discuss your feedback.

Statistics Median Online Tutor (p Nitin)

The median of a finite list of s can be found by arranging all the observations from lowest value to highest value and picking the middle one. If there is an even of observations, then there is no single middle value; the median is then defined to be the mean of the two middle values.

-Source Wikipedia.

Statistics is the proper science of creating well-organized use of numerical data between the sets of individuals. We can study the statistics in web sites. The web sites are providing the lot of definitions and example problems.

The online tutoring for the median is much helpful to the students to learn. By the online tutoring, students will get the help very interactively and also students can learn from their place.

Example to statistics median online tutor:

Example 1 by statistics online tutor:

Calculate the median for the given set of values. 23, 22, 86, 38, 30, 92, 26.

Solution:

First we have to arrange the given set of values in ascending order.

The ordered list of given set of values is 22,23,26,30,38,86,92.

Here, the given set is odd. So we have to pick out the middle.

Therefore, the median of the set of values is 30.

Example 2 by statistics online tutor:

Calculate the median for the given set of values. 48, 97, 55, 82, 27, 13, 14.

Solution:

First we have to arrange the given set in ascending order.

The ordered list of given set of values is 13,14,27,48,55,82,97

Here, the given set is odd. So we have to pick out the middle.

Therefore, the median of the given set of values is 48.

Example 3 by statistics online tutor:

How to find the median for the given set of values. 13, 78, 44, 61, 60, 77, 58, 20

Solution:

First we have to arrange the given set of values in ascending order.

The ordered list of given set of values is 13,20,44,58,60,61,77,78.

Here, the given set of values is even.

So we have to pick out the middle of two s and we have to solve average or mean of those values.

Therefore, the median of the set of values is '(58+60)/2' = 59.

Example 4 by statistics online tutor:

Calculate the median for the given set of values. 100, 26, 3, 80, 3, 48, 36, 11.

Solution:

First we have to arrange the given set of values in ascending order.

The ordered list of given set of values is 3, 3, 11, 26, 36 ,48, 80, 100.

Here, the given set is even.

So we have to pick out the middle of two s and we have to Solve average or mean of those values.

Therefore, the median of the set of values is '(26+36)/2' = 31.

Practice problems to statistics median online tutoring:

Problem 1 by statistics online tutor:

Calculate median for the given set of values. 95, 95, 4, 18, 79, 66, 97.

Solution: The median of the set of values is 79.

Problem 2 in online tutoring:

Calculate the median for the given set of values. 1, 71, 21, 67, 95, 3, 97, 6

Solution: The median of the given set of values is 44.

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Solve Learning Statistics (p Nitin)

In math, Statistics is the formal science of making effective use of numerical data connecting to groups of individuals or experiments. It contracts with all features of this, including not only the group, analysis and interpretation of such data, but also the planning of the group of data. Now we will discuss about how to solve and learning the statistics with some of the statistics example problems. (Source: Wikipedia).

Learning statistics:

In math, the statistics is very useful and important factor.
In statistics, the mean, median, range, and mode these are mostly used to learning the statistics.
Now we will learn the statistics with some of the examples.
Mean:

In solve learning statistics; the mean is detail about the middling of the given numbers.
So now we will find out the adding of the given numbers and divided by the total numbers of the given numbers.
Additions of the given numbers are 11, 25, 87.
= 11+25+87=123.

At this second divided by 3, because (Notes: 3 is the total given numbers whole)
= 123 / 3= 41.

Median:

In solve learning statistics; the median is described about the midpoint value of the given series.
The number series are 11, 25, 87.
The numbers in the middle of point of the above series are 25.
Therefore 25 is the median of the given series.

Mode and range in solve learning statistics:

Example of solve learning statistics:

Mode:

In solve learning statistics, the mode is prove again value of the specified number series; the particular value is shows again continually.
This is known as mode in solving statistics.
Example of mode: 11, 25, 87.
In this series are the no values are shows repeated.
So in this series the mode is blank.
Range:

The differentiation between greatest value and the lowest value is well-known as the range of the series.
87-11=76.
Therefore 76 is the range of the series in this series.

This is known as solve learning statistics.

Problem 1

Solve the mean, median and range of the following numbers in statistics?

16,18,21,26,28,31,35.

Solution

The given numbers are 16,18,21,26,28,31,35.

Mean

Mean is the average of the given number. Calculate the average with the help of total value of the given sequence.

Sum of the given numbers are = 16+18+21+26+28+31+35.

= 175.

Total values are divided by 7 (7 is the total numbers) = '(175)/(7)'

= 25.

Median

Center element of the given series is a median.

The number series is 16,18,21,26,28,31,35.

The middle element of the above sequence is 26.

So median=26.

Range

Less the low value from the high value.

Range=35-16

=19.

Example 2

Find out the mean, median, mode and range of the below numbers in statistics?

21,26,28,31,34,37,42.

Solution

The given numbers are 21,26,28,31,34,37,42.

Mean

Mean is the average of the given number. Calculate the average with the help of total value of the given sequence.

Sum of the given numbers are = 21+26+28+31+34+37+42

= 219.

Total values are divided by 7 (7 is the total numbers) = '(219)/(7)'

= 31.2.

Median

Middle element of the given series is a median.

The number series is 21,26,28,31,34,37,42.

The middle element of the above sequence is 31.

So median= 31.

Mode

Mode is the duplication of the series. The given sequence no copy value. Therefore the mode is empty.

Range

Less the low value from the high value of the series.

Range =42-21

=21.

These are the examples are helps to study the statistics in tutoring time.

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Laser Light Technologies (p Nitin)

A laser beam is a narrow beam of electromagnetic radiation which monochromatic, powerful and directional in nature. The laser beams are found within the visible spectrum of light. The atoms are excited by the laser device in the lasing medium. The electrons of these excited atoms jump from lower orbit to a higher orbit, and then release photons,thus creating a laser beam. The the acronym for LASER is the Light Amplification due to Stimulated Emission of Radiation. Laser can be defined as a mechanism for emitting light or visible light by a laser mechanism called process of stimulated emission. This is called as the electromagnetic radiation. The emitted laser light is generally spatially coherent, narrow beam which can be manipulated with the adjustment or modification of position of lenses. In laser technology, a light source is usually denoted as a "coherent light" as it emits light of in-step waves which are identical in frequency, polarization and phase. The laser's coherent light beam differentiates it from the other light sources which emit different incoherent light beams of random phase which vary along with position and time. Generally, Laser light is a electromagnetic spectrum monochromatic light of narrow wavelength. There are certain type of lasers which emit a broad spectrum of light or different wavelengths of light simultaneously.

A laser consists of gain medium inside an optical cavity which is highly reflective. It also contains a means to supply energy to the gain medium. The gain medium is a material that has properties which allow it to amplify the light received by stimulated emission. In a simple form, it can be explained as a cavity consisting of two mirrors arranged in such a way that light is bounced back and forth each time it is passed through the gain medium. Of the two mirrors, one is the output coupler which is a partially transparent mirror emits the output laser beam.A specific wavelength light which passes through the gain medium is undergoes amplification in order to increase the power and then the surrounding mirrors let most of the light to pass through the gain medium. A part of the light which is between the mirrors i.e., within the cavity passes through the partially transparent mirror and escapes as light.

The different types of lasers are liquid, solid and gas. Gas lasers are used to excite the electrons in gases such as carbon dioxide, nitrogen, helium, neon and cadmium. Liquid lasers such as the dye laser, uses organic dye molecules in the liquid form so as to produce a wavelength of radiation which can be tuned. Solid lasers such as the ruby laser uses a precious stone in order to produce a beam of red colored light.

Laser beams are commonly used in CD and DVD devices, optical scanners, computer mouse, laser printers and pointers for projectors. They are also used to produce hologramson several things. Laser beams are widely used in industry for cutting and welding the metal and to do survey of land and to construct buildings. They are used in scientific research for the chemical analysis and laser spectroscopy. For Dental applications including the treatment of cavity, regeneration of the nerve and reshaping tissue of the gum. In medical procedures such as eye surgeries, cancer treatments and heart surgery as well as in cosmetic procedures also laser beams are used.

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Extraction of Crude Metal from Concentrated Ore (p Nitin)

The process of extraction of metal from the concentrated ore depends upon the nature of the ore as well as the nature of the impurities present in the ore.Before the concentrated ore is subjected to final metallurgical operations in order to get the metal in the free state,the preliminary chemical treatment may be necessary.

The objective of the preliminary chemical treatment in extraction of crude metal from concentrated ore is :

(a)to get rid of impurities which would cause difficulties in the later stages; and

(b)to convert the ore into oxide of the metal.

The processes of extraction of crude metal from concentrated ore involved for preliminary treatment are Calcination and roasting.

(i)Calcination:It is a process of heating of ore in a limited supply of air below its melting point.The process involves:
-the removal of volatile impurities
-the removal of moisture
-the decomposition of any carbonate ore into oxide.
ZnCO3 gives ZnO + CO2
PbCO3 gives PbO + CO2
(ii)Roasting:It is the process of heating the ore in excess supply of air below its melting point.The process is employed when oxidation of the ore is required.As a result of roasting:
-moisture is driven away.
-volatile impurities are removed
-the impurities like sulphur,phosphorus,arsenic are removed as their oxides,
-the ore undergoes oxidation to form metal oxide.
2PbS + 3O2 gives 2PbO + 2SO2
Both calcination or roasting processes are carried out in a special type of furnace called reverberatory furnace.

extraction of crude metal from concentrated ore :- The reduction of ore

After the preliminary treatment in extraction of crude metal from concentrated ore,the ore may be subjected to reduction process by any one of the following methods depending upon its nature:

1.Smelting: reduction with carbon.In this process,the roasted or calcined ore is mixed with suitable quantity of coke or charcoal (which acts as reducing agent)and is heated to a high temperature above its melting point.During reduction,an additional reagent is also added to the ore to remove the impurities still present in the ore.The additiobal reagent is called flux.Flux combines with the impurities to form a fusible product called slag.

Flux + Impurities gives Slag
The selection of flux depends upon the nature of impurities.If impurities are acidic in nature,the flux is basic i.e. lime (CaO).On the other hand ,for basic impuritiies,an acidic flux,silica(SiO2 ) is used.
The process of reduction with carbon is carried out in a blast furnace.

2.Reduction with aluminium: Certain metal oxides such as Cr2O3 ,Mn3O4 are not easily reduced with carbon.In such cases,aluminium is used as reducing agent because it is more electropositive than chromium or manganese.The process of reduction of oxides with aluminium is called as aluminothermy.Some examples are:
Cr 2O3 + 2Al gives Al2O3 + 2Cr
3Mn3O4 + 8Al gives 4Al2O3 + 9Mn

3.Reduction with hydrogen or water gas.Hydrogen gas is very effective reducing agent.However its use is limited beacuse of its expensive nature.
WO3 + 3H2 gives W + 3H2O

4.Electrolytic reduction.The highly electropositive elements such as alkali metals ,alkaline earth metals and aluminium cannot be extracted by carbon reduction methods.They are extracted by the electrolysis of their fused salts.The process of extraction of metals by the use of electrolysis phenomenon is called electrometallurgy.

5.Leaching or hydrometallurgy.Some metals like gold or silver are extracted from their concentrated ores by leaching.They are dissolved in suitable reagents like acids or bases leaving behind soluble impurities.The metal is recovered from the solution by reprecipitation or crystallisation.For example,silver ore is leached with dilute solution of sodium cyanide.

These were the steps in extraction of crude metal from concentrated ore

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Laser Light Intensity (p Nitin)

Light amplification by stimulated emission of radiation (LASER or laser) is a mechanism for emitting electromagnetic radiation, typically light or visible light, via the process of stimulated emission. The emitted laser light is (usually) a spatially coherent, narrow low-divergence beam, that can be manipulated with lenses. In laser technology, "coherent light" denotes a light source that produces (emits) light of in-step waves of identical frequency, phase,and polarization. The laser's beam of coherent light differentiates it from light sources that emit incoherent light beams, of random phase varying with time and position. Laser light is generally a narrow-wavelength electromagnetic spectrum monochromatic light; yet, there are lasers that emit a broad spectrum of light, or emit different wavelengths of light simultaneously.

The word laser originally was the upper-case LASER, the acronym from Light Amplification by Stimulated Emission of Radiation, wherein light broadly denotes electromagnetic radiation of any frequency, not only the visible spectrum; hence infrared laser, ultraviolet laser, X-ray laser, et cetera. Because the microwave predecessor of the laser, the maser, was developed first, devices that emit microwave and radio frequencies are denoted "masers". In the early technical literature, especially in that of the Bell Telephone Laboratories researchers, the laser was also called optical maser, a currently uncommon term; moreover, since 1998, Bell Laboratories adopted the laser usage.Linguistically, the back-formation verb to lase means "to produce laser light" and "to apply laser light to" The word laser sometimes is used in an extended sense to describe a non-laser-light technology, e.g. a coherent-state atom source is an atom laser.

A laser consists of a gain medium inside a highly reflective optical cavity, as well as a means to supply energy to the gain medium. The gain medium is a material with properties that allow it to amplify light by stimulated emission. In its simplest form, a cavity consists of two mirrors arranged such that light bounces back and forth, each time passing through the gain medium. Typically one of the two mirrors, the output coupler, is partially transparent. The output laser beam is emitted through this mirror.

Light of a specific wavelength that passes through the gain medium is amplified (increases in power); the surrounding mirrors ensure that most of the light makes many passes through the gain medium, being amplified repeatedly. Part of the light that is between the mirrors (that is, within the cavity) passes through the partially transparent mirror and escapes as a beam of light.

The process of supplying the energy required for the amplification is called pumping. The energy is typically supplied as an electrical current or as light at a different wavelength. Such light may be provided by a flash lamp or perhaps another laser. Most practical lasers contain additional elements that affect properties such as the wavelength of the emitted light and the shape of the beam.

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Statistics Median Online Study (p Nitin)

Learning of statistics should be the formal science of generating the perfect help of mathematical data that are involving to collection of individuals. The term statistics should be either singular or else plural. The statistics learning includes the concepts of help with mean, median, range.

Statistics is the proper science of creating efficient use of arithmetical data relating to class of individuals. Statistics contract with all aspects including not only the collection and such data, but also the preparation of the group of data, in terms of the mean of experiments and surveys. Now we will learning the examples with the help of free statistics.

In probability theory and statistics, a median is described as the numeric value separating the higher half of a sample, a population, or a probability distribution, from the lower half.

Online:

In general, "online" indicates a state of connectivity, while "offline" indicates a disconnected state. In common usage, "online" often refers to the Internet or the World Wide Web.

(Source - Wikipedia.)

Formula for statistics median online study:

Formula for median of odd numbers:

The following formula which is used to find the median of the given odd number.

Median = ((n + 1)) / 2 .

n = total number of values in the collection.

Formula for median of even numbers:

The following formula which is used to calculate the median of the even numbers.

Median = (n/2) and (n / 2) + 1 .

n = total number of values in the collection.

Example problems for statistics median online study:

Statistics median online study - Example 1

1) Find the Median of 7, 3 and 8.

Solution:

Step 1: Given 7, 3 and 8.

Step 2: Arrange the number in ascending order 3, 7 and 8.

Step 3: total number in the set n = 3

Step 4: apply n =3 in the formula.

= ((n + 1)) / 2 .

= ((3 + 1)) / 2

= 4/2

= 2 nd value

The middle number is 7, so the median is 7

Answer is: 7.

Statistics median online study - Example 2

2) Find the median of 14, 13 and 12

Solution:

Step 1: Given 14, 13 and 12.

Step 2: Arrange the number in ascending order 12, 13 and 14.

Step 3: total number in the set n = 3

Step 4: apply n =3 in the formula.

= ((n + 1)) / 2 .

= ((3 + 1)) / 2

= 4/2

= 2 nd value

The middle number is 13, so the median is 13

Answer is: 13.

Statistics median online study - Example 3

3) Find the median of 12, 14, 13, 15, 11 and 16

Solution:

Step 1: Given 12, 14, 13, 15, 11 and 16

Step 2: Arrange the number in ascending order 11, 12, 13, 14, 15 and 16

Step 3: total number in the set n = 6

Step 4: apply n = 6 in the formula.

= (n/2) and (n/2) + 1 .

= 6/2 and (6 / 2) + 1

= 3 and 4

= 3rd and 4th value

Step 5: So, the median is

= ((13 + 14)) /2

= 27/2

= 13.5

Answer is: 13.5

Statistics median online study - Example 4

4) Find the median of 5, 11, 9, 6, 7, 10, 15 and 16

Solution:

Step 1: Given 5, 11, 9, 6, 7, 10, 15 and 16

Step 2: Arrange the number in ascending order 5, 6, 7, 9, 10, 11, 15 and 16

Step 3: total number in the set n = 8

Step 4: apply n = 8 in the formula.

= (n / 2) and (n/2) + 1 .

= 8/2 and (8 / 2) + 1

= 4 and 5

= 4th and 5th value

Step 5: So, the median is

= ((9 + 10)) /2

= 19/2

= 9.5

Answer is: 9.5

Practice problems for statistics median online study:

1) Find the median of 52, 47 and 65

Ans: 52

2) Find the median of 48, 55 and 80

Ans: 55

3) Find the median of 40, 24 and 30

Ans: 30

4) Find the median of 8, 10, 9, 11, 12 and 13

Ans: 10.5

5) Find the median of 5, 7, 6, 8, 10, 9, 11 and 12

Ans: 8.5

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Solving Statistics Homework Help (p Nitin)

In math, Statistics is the formal science of making effective use of numerical data connecting to groups of individuals or experiments. It contracts with all features of this, including not only the group, analysis and interpretation of such data, but also the planning of the group of data. Now we will discuss about solving statistics homework help with homework problems. (Source: Wikipedia).

Statistics is the proper science of creating valuable use of arithmetical data connecting to groups of individuals. Statistics compact with all features of this, with not only the group, analysis and interpretation of such data. Also the planning of the group of data, in terms of the design of surveys.Online tutoring is the most useful for study the statistics. We can learn with in an hour with the help of tutoring. This is the main advantage of the online tutoring.

In math, Statistics is the formal science of making effective use of numerical data connecting to groups of individuals or experiments. It contracts with all features of this, including not only the group, analysis and interpretation of such data, but also the planning of the group of data.

Solving Statistics:

In mathematics, the statistics is very priceless and major main part.
In statistics, the mean, median, range, and mode these are most important terms.
Now we will find out the homework problems with solving statistics help.
Mean:

In solving statistics homework help, the mean is one of the essential terms in statistics.
Now we will solve with the concept of mean with homework examples.
The mean is describes the average of the given numbers.

Therefore here we will get out the adding of the given numbers and divided by the total numbers of the given numbers.
The given numbers are 21, 40, 30.
Solution:

Additions of the given numbers are 21,40,30
21+40+30=91
The total is 91
Now divided by 3, for the reason that (Notes: 3 is the total given numbers)
= 91 /3=30.3.

So the mean value is 30.3.
Median:

In solving statistics homework help, the median is clarified about the medium value of the given series.
The number series are 21, 40, 30.
The numbers in the center point of point of the given series are 40.
Hence 40 is the median of the given numbers.

Mode and range in solving statistics homework help:

Example of solving statistics homework help:

Mode:

In solving statistics homework help, the mode is occurring more than one of the given number series; the particular value is shows again frequently.
Example of mode: 21, 4,89,4.
In this series 4 show again frequently.
So in this series the mode is 4.
Range:

The variation between very maximum value and the low value is known as the range in solving statistics.
The given values are 21, 4,89,4.
Here the highest value is 89, and then the lowest value is 4.
So the range value is difference between the 89 and 4.
The range value is 89-4=85.
So the range value is 85.
Homework 1:

The given series are 14,58,75,14.
Answer:

The mode is 14.
Homework 2:

The given numbers are 21,89,10,8,10.
Answer:

The mode is 10.

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