Tenth LectureSome Continuous Probability Distributions

Normal DistributionDefinition:The density of the normal random variable X,

with mean m and variance σ2 , is

Where And

Some properties of the normal curve:The mode, which is the point on the

horizontal axis the curve is a maximum, occurs at x= .m

The curve is symmetric about a vertical axis through the mean.

The curve has its points of inflection at is concave downward if and is

concave upward otherwise.

The normal curve approaches the horizontal axis asymptotically as we proceed in either direction away from the mean.

The total area under the curve and above the horizontal axis is equal to one.

The standard normal distributionDefinition:The distribution of a normal random variable

with mean 0 and variance 1 is called a standard normal distribution.

P(Z<-2.43)

Areas under the Normal CurveExample (1):Given a standard normal distribution, find the

area under the curve that lies To the right of z = 1.84=P(z > 1.84) = 1 - P(z

< 1.84) = 1 - 0.96712 = 0.03288 Between z= -1.97 and z= 0.86

Example (2)Given a standard normal distribution, find the

value of k such that P(Z > k)= .03015, andP(k < Z < -0.18)= .4197.

Using the normal curve in reverseExample (3)Given a normal distribution with = 40 m and

, find the value of x that has 45% of the area to the left, and14% of the area to the right.

Application of the Normal DistributionExample (4)A certain type of storage battery lasts, on

average, 3.0 years with a standard deviation of 0.5 years. Assuming that the battery lives are normally distributed find the probability that a given battery will last less than 2.3 years.

Normal Approximation to the Binomialthe relationship between the binomial and normal distribution is the binomial distribution is nicely approximated by the normal in practical problems. We now state a theorem that allows us to use areas under the normal curve to approximate binomial properties when n is sufficiently large.

Theorem: If X is a binomial random variable with mean =m np and variance s2=npq then the limiting form of the distribution of

as n is the standard normal distribution 11(z; 0,1).

X npZ

npq

To illustrate the normal approximation to the binomial distribution, we first draw the histogram for b(x; 10,0.5) and then superimpose the particular normal curve having the same mean and variance as the binomial variable X. Hence we draw a normal curve withm = np = (10)(0.5) = 5, and s2 = npq = (10)(0.5)(0.5) = 2.5.

P(X = 4) = b(4; 10,0.5) = 0.2051which is approximately equal to the area of the shaded region under the normal curve between the two ordinates x1 = 3.5 and x2 = 4.5 . Converting to z values, we have:

and

1

3.5 50.95

2.5z

2

4.5 50.32

2.5z

( 4) (4;10,0.5) ( 0.95 0.32)

( 0.95 0.32) ( 0.32) ( 0.95)

= 0.3745 0.1711

=0.2034

P X b P Z

P Z P Z P Z

This agrees very closely with the exact value of

( 4) 0.2051 P X

Definition:Let X be a binomial random variable with parameters n and p. Then X has approximation to approximately a normal distribution with m = np and s2 = npq = np(l — p) and

and the approximation will be good if np and n(1 —p) are greater than or equal to 5.

0

( ) ( ; , )

0.5

0.5 ( )

x

k

P X x b k n p

area under normal curve to the left of x

x npP Z

npq

Example (5):The probability that a patient recovers from a rare blood disease is 0.4. If 100 people are known to have contracted this disease, what is the probability that less than 30 survive?Solution :

Let the binomial variable X represent the number of patients that survive. Since n = 100, we should obtain fairly accurate results using the normal-curve approximation withm = np=(100)(0.4)=40, andwe have to find the area to the left of x =29.5.The z value corresponding to 29.5 is

Hence:

100 0.4 0.6 4.899npq

29.5 402.14

4.899z

( 30) (30;100,0.4) ( 2.14)= 0.0162P X b P Z

Example (6):A multiple-choice quiz has 200 questions each with 4 possible answers of which only 1 is the correct answer. What is the probability that sheer guesswork yields from 25 to 30 correct answers for 80 of the 200 problems about which the studenthas no knowledge?Solution: The probability of a correct answer for each of the 80 questions is p = 1/4. If X represents the number of correct answers due to guesswork, then

Using the normal-curve approximation with m =np=(80)(0.25)=20 , and

80 0.25 0.75 3.873npq

3014

25

(25 30) ( ;80, )x

P X b x

we need the area between x1 = 24.5 and x2 = 30.5. The corresponding z values are

and

The probability of correctly guessing from 25 to 30 questions is given by

2

30.5 202.71

3.873z

1

24.5 201.16

3.873z

(25 30) (1.16 2.71)

= ( 2.71) ( 1.16)

= 0.9966 - 0.8770

= 0.1196

P X P Z

P Z P Z

Chi-Squared Distribution:

F-Distribution:

t-Distribution:

Good luck