Unit 4
Statistical techniques2
Probability is the study of chances. Probability is the measurement of the degree of uncertainty and therefore, of certainty of the occurrence of events.
A probability space is a threetuple (S, F, P) in which the three components are
Basic definitions
1. Exhaustive events The set of all possible outcomes of an experiment is called an exhaustive event of sample space.
Example
1. If we toss a coin then the sample space is
S = {H, T}, where H and T denote head and tail respectively and n(S) = 2
2. If a coin is tossed thrice or three coins are tossed simultaneously, then the sample space is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and n(S) = 8.
3. If a coin is tossed 4 times or four coins are tossed simultaneously then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT,
THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} and n(S) = 16.
Each outcome is called a sample point.
Example If a die is thrown twice, then getting (1, 1) or (1, 2) or (1, 3) or…or (6, 6) is a sample point.
2. Mutually exclusive events When the occurrence of one event excludes the occurrence of the other than these two events said to be mutually exclusive
3. Equally likely Two events are said to be equally likely if one of them cannot occur in preference to the others.
Random experiment
An experiment in which all the possible outcomes are known in advance but we cannot predict which of them will occur when we experiment.
Example‘Throwing a die’ and ‘Drawing a card from a wellshuffled pack of 52 playing cards ‘are examples of a random experiment
Event
A set of one or more possible outcomes of an experiment constitutes what is known as an event. Thus, an event can be defined as a subset of the sample space
Favourable cases
The cases which favour the happening of an event are called favourable cases
ExampleFor the event of getting an even number in throwing a die, the number of favourable cases is 3 and the event, in this case, is {2, 4, 6}.
Odds in favour of an event and odds against an event
If the number of favourable cases is ‘m’ and the number or not favourable cases are ‘n’.
Then
1. Odds in favour of the event = m/n
2. Odds against the event = n/m
The classical definition of probability
Suppose there are ‘n’ exhaustive cases in a random experiment that is equally likely and mutually exclusive.
Let ‘m’ cases are favourable for the happening of an event A, then the probability of happening event A can be defined as
The probability of nonhappening of the event A is defined as
Note Always remember that the probability of any events lies between 0 and 1.
Expected value
Let are the probabilities of events and respectively. Then the expected value can be defined as
Example: A bag contains 7 red and 8 black balls then find the probability of getting a red ball.
Sol.
Here total cases = 7 + 8 = 15
According to the definition of probability,
So that, here favourable cases red balls = 7
Then,
Addition and multiplication law of probability
Addition law
If are the probabilities of mutually exclusive events, then the probability P, that any of these events will happen is given by
Note
If two events A and B are not mutually exclusive then the probability of the event that either A or B or both will happen is given by
Example: A box contains 4 white and 2 black balls and a second box contains three balls of each colour. Now a bag is selected at random and a ball is drawn randomly from the chosen box. Then what will be the probability that the ball is white?
Sol.
Here we have two mutually exclusive cases
1. The first bag is chosen
2. The second bag is chosen
The chance of choosing the first bag is 1/2. And if this bag is chosen then the probability of drawing a white ball is 4/6.
So that the probability of drawing a white ball from the first bag is
And the probability of drawing a white ball from the second bag is
Here the events are mutually exclusive, then the required probability is
Example25 lottery tickets are marked with the first 25 numerals. A ticket is drawn at random.
Find the probability that it is a multiple of 5 or 7.
Sol:
Let A be the event that the drawn ticket bears a number multiple of 5 and B be the event that it bears a number multiple of 7.
So that
A = {5, 10, 15, 20, 25}
B = {7, 14, 21}
Here, as A B = ,
A and B are mutually exclusive
Then,
Multiplication law
For two events A and B
Here is called conditional probability of B given that A has already happened.
Now
If A and B are two independent events, then
Because in the case of independent events
Example: A bag contains 9 balls, two of which are red three blue, and four black.
Three balls are drawn randomly. What is the probability that
1. The three balls are of different colours
2. The three balls are of the same colours.
Sol.
1. Three balls will be of a different colour if one ball is red, one blue and one black ball are drawn
Then the probability will be
2. Three balls will be of the same colour if one ball is red, one blue and one black ball are drawn
Then the probability will be
Example: A die is rolled. If the outcome is a number greater than three. What is the probability that it is a prime number?
Sol.
The sample space is S = {1, 2, 3, 4, 5, 6}
Let A be the event that an outcome is a number that is greater than three and B be the event that it is a prime.
So that
A = {4, 5, 6} and B = {2, 3, 5} and hence
P(A) = 3/6, P(B) = 3/6 and
Now the required probability
Example: Two cards are drawn from a pack of playing cards in succession with the replacement of the first card. Find the probability that both are the cards of heart.
Sol.
Let A be the event that the first card drawn is a heart and B be the event that the second card is a heart card.
As the cards are drawn with replacement,
Here A and B are independent and the required probability will be
Example: Two male and female candidates appear in an interview for two positions in the same post. The probability that the male candidate is selected is 1/7 and the female candidate selected is 1/5.
What is the probability that
1. Both of them will be selected
2. Only one of them will be selected
3. None of them will be selected.
Sol.
Here, P (male’s selection) = 1/7
And
P (female’s selection) = 1/5
Then
3.
Example: A can hit a target 3 times in 5 shots, B 2 times in 5 shots, and C 3 times in 4 shots. All of them fire one shot each simultaneously at the target.
What is the probability that
1. Two shots hit
2. At least two shots hit
Sol.
1. Now probability that 2 shots hit the target
2.
Probability of at least two shots hitting the target
Let A and B be two events of a sample space Sand let . Then the conditional probability of the event A, given B, denoted byis defined by –
Theorem: If the events A and B defined on a sample space S of a random experiment are independent, then
Example1:A factory has two machines A and B making 60% and 40% respectively of the total production. Machine A produces 3% defective items, and B produces 5% defective items. Find the probability that a given defective part came from A.
SOLUTION: We consider the following events:
A: Selected item comes from A.
B: Selected item comes from B.
D: Selected item is defective.
We are looking for a . We know:
Now,
So we need
Since D is the union of the mutually exclusive events and (the entire sample space is the union of the mutually exclusive events A and B)
Example: Two fair dice are rolled, 1 red, and 1 blue. The Sample Space is
S = {(1, 1),(1, 2), . . . ,(1, 6), . . . ,(6, 6)}.Total 36 outcomes, all equally likely (here (2, 3) denotes the outcome where the red die show 2 and the blue one shows 3).
(a) Consider the following events:
A: Red die shows 6.
B: Blue die shows 6.
Find , and .
Solution:
NOTE:so for this example. This is not surprising  we expect A to occur in of cases. In of these cases i.e. in of all cases, we expect B to also occur.
(b) Consider the following events:
C: Total Score is 10.
D: Red die shows an even number.
Find , and .
Solution:
NOTE:so, .
Why does multiplication not apply here as in part (a)?
ANSWER: Suppose C occurs: so the outcome is either (4, 6), (5, 5), or (6, 4). In two of these cases, namely (4, 6) and (6, 4), the event D also occurs. Thus
Although , the probability that D occurs given that C occurs is .
We write , and call the conditional probability of D given C.
NOTE: In the above example
Example: Three urns contain 6 red, 4 black; 4 red, 6 black; 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball is drawn is red find the probability that it is drawn from the first urn.
Solution:
: The ball is drawn from urn I.
: The ball is drawn from urn II.
: The ball is drawn from urn III.
R: The ball is red.
We have to find
Since the three urns are equally likely to be selected
Also,
From (i), we have
Example: A bag contains 12 pens of which 4 are defective. Three pens are picked at random from the bag one after the other.
Then find the probability that all three are nondefective.
Sol. here the probability of the first which will be nondefective = 8/12
By the multiplication theorem of probability,
If we draw pens one after the other then the required probability will be
Example: The probability of A hits the target is 1 / 4 and the probability that B hits the target is 2/ 5. If both shoot the target then find the probability that at least one of them hits the target.
Sol.
Here it is given that
Now we have to find
Both two events are independent. So that
4.3. Bayes’ theorem
Bayes' rule
If , are mutually exclusive events with of a random experiment than for any arbitrary event of the sample space of the above experiment with , we have
(for )
Example1: An urn contains 3 white and 4 red balls and an urn lI contains 5 white and 6 red balls. One ball is drawn at random from one of the urns and is found to be white. Find the probability that it was drawn from urn 1.
Solution: Let : the ball is drawn from urn I
: the ball is drawn from urn II
: the ball is white.
We have to find
By Bayes Theorem
... (1)
Since two urns are equally likely to be selected, (a white ball is drawn from urn )
(a white ball is drawn from urn II)
From(1),
Example2: Three urns contain 6 red, 4 black, 4 red, 6 black; 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. lf the ball is drawn is red find the probability that it is drawn from the first urn.
Solution: Let : the ball is drawn from urn 1.
: the ball is drawn from urn lI.
: the ball is drawn from urn 111.
: the ball is red.
We have to find .
By Baye’s Theorem,
Since the three urns are equally likely to be selected
Also (a red ball is drawn from urn )
(R/) (a red ball is drawn from urn II)
(a red ball is drawn from urn III)
From (1), we have
Example3: ln a bolt factory machines and manufacturerespectively 25%, 35% and 40% of the total. lf their output 5, 4, and 2 percent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine B.?
Solution: bolt is manufactured by machine
: bolt is manufactured by machine
: bolt is manufactured by machine
The probability of drawing a defective bolt manufactured by machine is (D/A)
Similarly, (D/B) and (D/C)
By Baye’s theorem
A random variable is said to be discrete if it has either a finite or a countable number of values
The number of students present each day in a class during an academic session is an example of a discrete random variable as the number cannot take a fractional value.
Probability mass function
Let X be an r.v. which takes the values and let P[X = ] = p(. This function p(xi), i =1,2, … defined for the values assumed by X is called probability mass function of X satisfying p(xi) ≥0 and
Example: A random variable x has the following probability distribution
Then find
1. Value of c.
2. P[X≤3]
3. P[1 < X <4]
Sol.
We know that for the given probability distribution
So that
2.
3.
Independent random variables
Two discrete random variables X and Y are said to be independent only if
Note Two events are independent only if
Example: Two discrete random variables X and Y have
And
P[X = 1, Y = 1] = 5/9.
Check whether X and Y are independent or not?
Sol.
First, we write the given distribution In tabular form
X/Y  0  1  P(x) 
0  2/9  1/9  3/9 
1  1/9  5/9  6/9 
P(y)  3/9  6/9  1 
Now
But
So that
Hence X and Y are not independent.
Continuous Random Variables:
A continuous random variable is a random variable where the data can take infinitely many values. For example, a random variable measuring the time taken for something to be done is continuous since there is an infinite number of possible times that can be taken.
A continuous random variable is called by a probability density function p (x), with given properties: p (x) ≥ 0 and the area between the xaxis & the curve is 1: ... standard deviation of a variable Random is defined by σ x = √Variance (x).
∫∞∞ p(x) dx = 1.
2. The expected value E(x) of a discrete variable is known as:
E(x) = Σi=1n xi pi
3. The expected value E(x) of a continuous variable is called:
E(x) = ∫∞∞ x p(x) dx
4. The Variance(x) of a random variable is known as Variance(x) = E[(x  E(x)2].
5. 2 random variable x and y are independent if E[xy] = E(x)E(y).
6. The standard deviation of a random variable is known as σx = √Variance(x).
7. Given the value of standard error is used in its place of a standard deviation when denoting the sample mean.
σmean = σx / √n
8. If x is a normal random variable with limitsμ and σ2 (spread = σ), mark in symbols: x ˜ N(μ, σ2).
9. The sample variance of x1, x2, ..., xn is given by
sx2 = 

10. If x1, x2, ..., xn are explanations since a random sample, the sample standard deviation s is known as the square root of variance:
sx =  √ 

11. Sample Covariance of x1, x2, ..., xn is known
sxy = 

12. A random vector is a column vector of a random variable.
v = (x1 ... xn)T
13. The expected value of Random vector E(v) is known by a vector of the expected value of the component.
If v = (x1 ... xn)T
E(v) = [E(x1) ... E(xn)]T
14. Covariance of matrix Covariance(v) of a random vector is the matrix of variances and Covariance of the component.
If v = (x1 ... xn)T, the ijth component of the Covariance(v) is sij
Properties
Starting from properties 1 to 7, c is a constant; x and y are random variables.
From given properties 8 to 12, w and v is a random vector; b is a continuous vector; A is a continuous matrix.
8. E(v + w) = E(v) + E(w)
9. E(b) = b
10. E(Av) = A E(v)
11. Covariance(v + b) = Covariance(v)
12. Covariance(Av) = A Covariance(v) AT
Let a random variable X has a probability distribution which assumes the values say with their associated probabilities then the mathematical expectation can be defined as
The expected value of a random variable X is written as E(X).
The expected value for a continuous random variable is
OR
The mean value (μ) of the probability distribution of a variate X is commonly known as its expectation current is denoted by E (X). If f(x) is the probability density function of the variate X, then
(discrete distribution)
(continuous distribution)
In general expectation of any function is given by
(discrete distribution)
(continuous distribution)
(2) Variance offer distribution is given by
(discrete distribution)
(continuous distribution)
Where is the standard deviation of the distribution.
(3) The rth moment about the mean (denoted by is defined by
(discrete function)
(continuous function)
(4) Mean deviation from the mean is given by
(discrete distribution)
(continuous distribution)
Example. In a lottery, m tickets are drawn at a time out of tickets numbered from 1 to n. Find the expected value of the sum of the numbers on the tickets drawn.
Solution. Let be the variables representing the numbers on the first, second,…nth ticket. The probability of drawing a ticket out of n ticket spelling in each case 1/n, we have
Therefore the expected value of the sum of the numbers on the tickets drawn
Example. X is a continuous random variable with probability density function given by
Find k and mean value of X.
Solution. Since the total probability is unity.
Mean of X =
Example. The frequency distribution of a measurable characteristic varying between 0 and 2 is as under
Calculate two standard deviations and also the mean deviation about the mean.
Solution. Total frequency N =
(about the origin)=
(about the origin)=
Hence,
i.e., standard deviation
Mean derivation about the mean
Example: If a random variable X has the following probability distribution in the tabular form then what will be the expected value of X.
X  0  1  2 
P(x)  1/4  1/2  1/4 
Sol.
We know that
So that
Example: Find the expectations of the number of an unbiased die when thrown.
Sol. Let X be a random variable that represents the number on a die when thrown.
X can take the values
1, 2, 3, 4, 5, 6
With
P[X = 1] = P[X = 2] = P[X = 3] = P[X = 4] = P[X = 5] = P[X = 6] = 1/6
The distribution table will be
X  1  2  3  4  5  6 
p(x)  1/6  1/6  1/6  1/6  1/6  1/6 
Hence the expectation of number on the die thrown is
So that
The variance of a sum
One of the applications of covariance is finding the variance of a sum of several random variables. In particular, if Z = X + Y, then
Var (Z) =Cov (Z,Z)
More generally, for a, bR we conclude
Variance
Consider two random variables X and Y with the following PMFs
(3.3)
(3.4)
Note that EX =EY = 0. Although both random variables have the same mean value, their distribution is completely different. Y is always equal to its mean of 0, while X is IDA 100 or 100, quite far from its mean value. The variance is a measure of how spread out the distribution of a random variable is. Here the variance of Y is quite small since its distribution is concentrated value. Why the variance of X will be larger since its distribution is more spread out.
The variance of a random variable X with mean , is defined as
By definition, the variance of X is the average value of Since ≥0, the variance is always larger than or equal to zero. A large value of the variance means that is often large, so X often X value far from its mean. This means that the distribution is very spread out. on the other hand, low variance means that the distribution is concentrated around its average.
Note that if we did not square the difference between X and it means the result would be zero. That is
X is sometimes below its average and sometimes above its average. Thus is sometimes negative and sometimes positive but on average it is zero.
To compute , note that we need to find the expected value of , so we can use LOTUS. In particular, we can write
For example, for X and Y defined in equations 3.3 and 3.4 we have
As we expect, X has a very large variance while Var (Y) = 0
Note that Var (X) has a different unit than X. For example, if X is measured in metres then Var(X) is in .to solve this issue we define another measure called the standard deviation usually shown as which is simply the square root of variance.
The standard deviation of a random variable X is defined as
The standard deviation of X has the same unit as X. For X and Y defined in equations 3.3 and 3.4 we have
Here is a useful formula for computing the variance.
The computational formula for the variance
To prove it note that
Note that for a given random variable X, is just a constant real number. Thus so we have
Equation 3.5 is equally easier to work with compared to . To use this equation we can find using LOTUS.
And then subtract to obtain the variance.
Example. I roll a fair die and let X be the resulting number. Find E(X), Var(X), and
Solution. We have and for k = 1,2,…,6. Thus we have
Thus,
Theorem
For random variable X and real number a and b
Proof. If
From equation 3.6, we conclude that, for standard deviation, . We mentioned that variance is NOT a linear operation. But there is a very important case, in which variance behaves like a linear operation and that is when we look at the sum of independent random variables,
Theorem
If are independent random variables and , then
Example. If Binomial (n, p) find Var (X).
Solution. We know that we can write a Binomial (n, p) random variable as the sum of n independent Bernoulli (p) random variable, i.e.
If Bernoulli (p) then its variance is
Problem. If , find Var (X).
Solution. We already know , thus Var (X). You can find directly using LOTUS, however, it is a little easier to find E [X (X1)] first. In particular using LOTUS, we have
S
So we have . Thus, and we conclude
BINOMIAL DISTRIBUTION
To find the probability of the happening of an event once, twice, thrice,…r times ….exactly in n trails.
Let the probability of the happening of an event A in one trial be p and its probability of not happening to be 1 – p – q.
We assume that there are n trials and the happening of the event A is r times and it's not happening is n – r times.
This may be shown as follows
AA……A
r times n – r times (1)
A indicates it's happening its failure and P (A) =p and P (
We see that (1) has the probability
pp…p qq….q=
r times nr times (2)
Clearly (1) is merely one order of arranging r A’S.
The probability of (1) =Number of different arrangements of r A’s and (nr)’s
The number of different arrangements of r A’s and (nr)’s
Probability of the happening of event r times =
If r = 0, probability of happening of an event 0 times
If r = 1, probability of happening of an event 1 times
If r = 2, probability of happening of an event 2 times
If r = 3, probability of happening of an event 3 times and so on.
These terms are the successive terms in the expansion of
Hence it is called Binomial Distribution.
Example. If on average one ship in every ten is wrecked. Find the probability that out of 5 ships expected to arrive, 4 at least we will arrive safely.
Solution. Out of 10 ships, one ship is wrecked.
I.e. nine ships out of 10 ships are safe, P (safety) =
P (at least 4 ships out of 5 are safe) = P (4 or 5) = P (4) + P(5)
Example. The overall percentage of failures in a certain examination is 20. if 6 candidates appear in the examination what is the probability that at least five pass the examination?
Solution. Probability of failures = 20%
Probability of (P) =
Probability of at least 5 passes = P(5 or 6)
Example. The probability that a man aged 60 will live to be 70 is 0.65. what is the probability that out of 10 men, now 60, at least seven will live to be 70?
Solution. The probability that a man aged 60 will live to be 70
Number of men= n = 10
Probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)
= P (7)+ P(8)+ P(9) + P(10) =
Example. assuming that 20% of the population of a city are literate so that the chance of an individual being literate is and assuming that a hundred investigators each take 10 individuals to see whether they are illiterate, how many investigators would you expect to report 3 or less were literate.
Solution.
Required number of investigators = 0.879126118× 100 =87.9126118
= 88 approximate
Example: A die is thrown 8 times then find the probability that 3 will show
1. Exactly 2 times
2. At least 7 times
3. At least once
Sol.
As we know that
Then
1. Probability of getting 3 exactly 2 times will be
2. Probability of getting 3 at least 7 or 8 times will be
3. Probability of getting 3 at least once or (1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 times)
Mean of the binomial distribution
Successors r  Frequency f  rf 
0  0  
1  
2  n(n1)  
3  
…..  ……  …. 
N 
Since
STANDARD DEVIATION OF BINOMIAL DISTRIBUTION
Successors r  Frequency f  
0  0  
1  
2  2n(n1)  
3  
…..  ……  …. 
N 
We know that (1)
r is the deviation of items (successes) from 0.
Putting these values in (1) we have
Hence for the binomial distribution, Mean
Example. A die is tossed thrice. Success is getting 1 or 6 on a TOSS. Find the mean and variance of the number of successes.
Solution.
Example: Find the mean and variance of a binomial distribution with p = 1/4 and n = 10.
Sol.
Here
Mean = np =
Variance = npq =
RECURRENCE RELATION FOR THE BINOMIAL DISTRIBUTION
By Binomial Distribution
On dividing (2) by (1), we get
Poisson Distribution:
Poisson distribution is a particular limiting form of the Binomial distribution when p (or q) is very small and n is large enough.
Poisson distribution is
where m is the mean of the distribution.
Proof. In Binomial Distribution
Taking limits when n tends to infinity
MEAN OF POISSON DISTRIBUTION
Success r  Frequency f  f.r 
0  0  
1  
2  
3  
…  …  … 
r  
…  …  … 
STANDARD DEVIATION OF POISSON DISTRIBUTION
Successive r  Frequency f  Product rf  Product 
0  0  0  
1  
2  
3  
…….  ……..  ……..  …….. 
r  
……..  …….  ……..  ……. 
Hence mean and variance of a Poisson distribution is equal to m. Similarly, we can obtain,
MEAN DEVIATION
Show that in a Poisson distribution with unit mean, and the mean deviation about the mean is 2/e times the standard deviation.
Solution. But mean = 1 i.e. m =1 and S.D. =
r  P (r)  r1  P(r)r1 
0  1  
1  0  0  
2  1  
3  2  
4  3  
…..  …..  …..  ….. 
r  r1 
Mean Deviation =
MOMENT GENERATING FUNCTION OF POISSON DISTRIBUTION
Solution.
Let be the moment generating function then
CUMULANTS
The cumulant generating function is given by
Now cumulant =coefficient of in K (t) = m
i.e. , where r = 1,2,3,…
Mean =
RECURRENCE FORMULA FOR POISSON DISTRIBUTION
SOLUTION. By Poisson distribution
On dividing (2) by (1) we get
Example. Assume that the probability of an individual coal miner being killed in a mine accident during a year is . Use appropriate statistical distribution to calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year.
Solution.
Example. Suppose 3% of bolts made by a machine are defective, the defects occurring at random during production. If bolts are packaged 50 per box, find
(a) Exact probability and
(b) Poisson approximation to it, that a given box will contain 5 defectives.
Solution.
(a) Hence the probability for 5 defectives bolts in a lot of 50.
(b) To get Poisson approximation m = np =
Required Poisson approximation=
Example. In a certain factory producing cycle tyres, there is a small chance of 1 in 500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the approximate number of lots containing no defective, one defective, and two defective tyres, respectively, in a consignment of 10,000 lots.
Solution.
S.No.  Probability of defective  Number of lots containing defective 
1.  
2.  
3. 
Normal Distribution
The normal distribution is a continuous distribution. It is derived as the limiting form of the Binomial distribution for large values of n and p and q are not very small.
The normal distribution is given by the equation
(1)
Where = mean, = standard deviation, =3.14159…e=2.71828…
On substitution in (1) we get (2)
Here mean = 0, standard deviation = 1
(2) is known as the standard form of normal distribution.
Graph of a normal probability function
The curve looks like a bellshaped curve. The top of the bell is exactly above the mean.
If the value of standard deviation is large then the curve tends to flatten out and for small standard deviation, it has a sharp peak.
This is one of the most important probability distributions in statistical analysis.
MEAN FOR NORMAL DISTRIBUTION
Mean [Putting
STANDARD DEVIATION FOR NORMAL DISTRIBUTION
Put,
MEDIAN OF THE NORMAL DISTRIBUTION
If a is the median then it divides the total area into two equal halves so that
Where,
Suppose mean, then
Thus,
Similarly, when mean, we have a =
Thus, median = mean =
MEA DEVIATION ABOUT THE MEAN
Mean deviation
MODE OF THE NORMAL DISTRIBUTION
We know that mode is the value of the variate x for which f (x) is maximum. Thus by differential calculus f (x) is maximum if and
Where,
Thus, mode is and model ordinate =
NORMAL CURVE
Let us show binomial distribution graphically. The probabilities of heads in 1 toss are
. it is shown in the given figure.
If the variates (head here) are treated as if they were continuous, the required probability curve will be normal as shown in the above figure by dotted lines.
Properties of the normal curve
AREA UNDER THE NORMAL CURVE
By taking the standard normal curve is formed.
The total area under this curve is 1. The area under the curve is divided into two equal parts by z = 0. The lefthand side area and righthand side area to z = 0 are 0.5. The area between the ordinate z = 0.
Example. On a final examination in mathematics, the mean was 72, and the standard deviation was 15. Determine the standard scores of students receiving grades.
(a) 60
(b) 93
(c) 72
Solution. (a)
(b)
(c)
Example. Find the area under the normal curve in each of the cases
(a) Z = 0 and z = 1.2
(b) Z = 0.68 and z = 0
(c) Z = 0.46 and z = 2.21
(d) Z = 0.81 and z = 1.94
(e) To the left of z = 0.6
(f) Right of z = 1.28
Solution. (a) Area between Z = 0 and z = 1.2 =0.3849
(b)Area between z = 0 and z = 0.68 = 0.2518
(c)Required area = (Area between z = 0 and z = 2.21) + (Area between z = 0 and z =0.46)\
= (Area between z = 0 and z = 2.21)+ (Area between z = 0 and z = 0.46)
=0.4865 + 0.1772 = 0.6637
(d)Required area = (Area between z = 0 and z = 1.+(Area between z = 0 and z = 0.81)
= 0.47380.2910=0.1828
(e) Required area = 0.5(Area between z = 0 and z = 0.6)
= 0.50.2257=0.2743
(f)Required area = (Area between z = 0 and z = 1.28)+0.5
= 0.3997+0.5
= 0.8997
Example. The mean inside diameter of a sample of 200 washers produced by a machine is 0.0502 cm and the standard deviation is 0.005 cm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 cm, otherwise, the washers are considered defective. Determine the percentage of defective washers produced by the machine assuming the diameters are normally distributed.
Solution.
Area for non – defective washers = Area between z = 1.2
And z = +1.2
=2 Area between z = 0 and z = 1.2
=2 (0.3849)0.7698=76.98%
Percentage of defective washers = 10076.98=23.02%
Example. A manufacturer knows from experience that the resistance of resistors he produces is normal with mean and standard deviation . What percentage of resistors will have resistance between 98 ohms and 102 ohms?
Solution. ,
Area between
= (Area between z = 0 and z = +1)
= 2 (Area between z = 0 and z = +1)=2 0.3413 = 0.6826
Percentage of resistors having resistance between 98 ohms and 102 ohms = 68.26
Example. In a normal distribution, 31% of the items are 45 and 8% are over 64. Find the mean and standard deviation of the distribution.
Solution. Let be the mean and the S.D.
If x = 45,
If x = 64,
The area between 0 and
[From the table, for the area 0.19, z = 0.496)
Area between z = 0 and z =
(from the table for area 0.42, z = 1.405)
Solving (1) and (2) we get
Example: The life of electric bulbs is normally distributed with a mean of 8 months and a standard deviation of 2 months.
If 5000 electric bulbs are issued how many bulbs should be expected to need replacement after 12 months?
[Given that P (z ≥ 2) = 0. 0228]
Sol.
Here mean (μ) = 8 and standard deviation = 2
Number of bulbs = 5000
Total months (X) = 12
We know that
Area (z ≥ 2) = 0.0228
Number of electric bulbs whose life is more than 12 months ( Z> 12)
= 5000 × 0.0228 = 114
Therefore replacement after 12 months = 5000 – 114 = 4886 electric bulbs.
Example:
1. If X then find the probability density function of X.
2. If X then find the probability density function of X.
Sol.
1. We are given X
Here
We know that
Then the p.d.f. will be
2. . We are given X
Here
We know that
Then the p.d.f. will be
Example: If a random variable X is normally distributed with mean 80 and standard deviation 5, then find
1. P[X > 95]
2. P[X < 72]
3. P [85 < X <97]
[Note use the table area under the normal curve]
Sol.
The standard normal variate is –
Now
1. X = 95,
So that
2. X = 72,
So that
3. X = 85,
X = 97,
So that
Example: In a company, the mean weight of 1000 employees is 60kg and the standard deviation is 16kg.
Find the number of employees having their weights
1. Less than 55kg.
2. More than 70kg.
3. Between 45kg and 65kg.
Sol. Suppose X be a normal variate = the weight of employees.
Here mean 60kg and S.D. = 16kg
X
Then we know that
We get from the data,
Now
1. For X = 55,
So that
2. For X = 70,
So that
3. For X = 45,
For X = 65,
Hence the number of employees having weights between 45kg and 65kg
References
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.
3. P. G. Hoel, S. C. Port, and C. J. Stone, Introduction to Probability Theory, Universal Book Stall.
4. S. Ross, A First Course in Probability, 6th Ed., Pearson Education India,2002.