Assume that women’s heights are normally distributed with a mean of 63.6 inches
ID: 3337394 • Letter: A
Question
Assume that women’s heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If one woman is randomly selected, find the probability that she has a height between 62.9 inches and 64.0 inches
In a study, 44% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 12 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health.
The amount of snow falling in a certain mountain region is normally distributed with a mean of 76 inches, and a standard deviation of 14 inches. What is the probability that the annual snowfall of a randomly picked year will be 78.8 inches or less?
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. In a random sample of 9000, approximately how many people will have IQs between 85 and 120?
A bank’s loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. Find P60, the score which separates the lower 60% from the top 40%.
A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If a Chevrolet Cavalier is randomly selected, find the probability that its rebuild time exceeds 8.7 hours. Assume normality.
Assume that women’s heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If one woman is randomly selected, find the probability that she has a height between 62.9 inches and 64.0 inches
In a study, 44% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 12 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health.
The amount of snow falling in a certain mountain region is normally distributed with a mean of 76 inches, and a standard deviation of 14 inches. What is the probability that the annual snowfall of a randomly picked year will be 78.8 inches or less?
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. In a random sample of 9000, approximately how many people will have IQs between 85 and 120?
A bank’s loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. Find P60, the score which separates the lower 60% from the top 40%.
A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If a Chevrolet Cavalier is randomly selected, find the probability that its rebuild time exceeds 8.7 hours. Assume normality.
Explanation / Answer
Q1.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 63.6
standard Deviation ( sd )= 2.5/ Sqrt ( 1 ) =2.5
sample size (n) = 1
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 62.9) = (62.9-63.6)/2.5/ Sqrt ( 1 )
= -0.7/2.5
= -0.28
= P ( Z <-0.28) From Standard Normal Table
= 0.3897
P(X < 64) = (64-63.6)/2.5/ Sqrt ( 1 )
= 0.4/2.5 = 0.16
= P ( Z <0.16) From Standard Normal Table
= 0.5636
P(62.9 < X < 64) = 0.5636-0.3897 = 0.1738
Q2.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 12 * 0.44
= 5.28
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 12 * 0.44 * 0.56
= 2.9568
III.
standard deviation = sqrt( variance ) = sqrt(2.9568)
=1.7195
a.
P( X = 3 ) = ( 12 3 ) * ( 0.44^3) * ( 1 - 0.44 )^9
= 0.1015
b.
P( X < = 3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 12 3 ) * 0.44^3 * ( 1- 0.44 ) ^9 + ( 12 2 ) * 0.44^2 * ( 1- 0.44 ) ^10 + ( 12 1 ) * 0.44^1 * ( 1- 0.44 ) ^11 + ( 12 0 ) * 0.44^0 * ( 1- 0.44 ) ^12
= 0.1502