1. In a student population, the height of women is normally distributed with mea
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Question
1. In a student population, the height of women is normally distributed with mean 63.5 inches and standard deviation 2.9 inches. (a) What percentage of women will have a height of greater than 64 inches? (b) Find the 20'th and 80'th percentiles of the height distribution for women. That is, find two numbers q20, 980 so that 20 percent of women are shorter than q20 and 80 percent of women are shorter than qso. (c) We want to choose a sample of n women so that, with probability of at least 90 percent, at least one woman in the sample is more than 72 inches tall. What is the smallest value of n E N with this property? (d) Assume that a sample of size n = 7 is chosen at random. What is the probability that the sample mean falls in the interval 63, 64?Explanation / Answer
1.
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 ~ N(0,1)
mean ( u ) = 63.5
standard Deviation ( sd )= 2.9
a. percenge of women will have height of greater than 64 inches
P(X > 64) = (64-63.5)/2.9
= 0.5/2.9 = 0.1724
= P ( Z >0.1724) From Standard Normal Table
= 0.4316 = 43.16%
b.
P ( Z < x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is -0.84
P( x-u/s.d < x - 63.5/2.9 ) = 0.2
That is, ( x - 63.5/2.9 ) = -0.84
--> x = -0.84 * 2.9 + 63.5 = 61.06
P ( Z > x ) = 0.8
Value of z to the cumulative probability of 0.8 from normal table is -0.84
P( x-u / (s.d) > x - 63.5/2.9) = 0.8
That is, ( x - 63.5/2.9) = -0.84
--> x = -0.84 * 2.9+63.5 = 61.06
c.
Z = (X-u)/(sd/sqrt(n)
P ( Z > x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is -1.28
more than 72 inches one woman atleast
-1.28 = (72-63.5)/(2.9)/sqrt (n)
sqrt (n) = -1.28*2.931 = - 3.751
n = 14.07 = 14
d.
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 of the sampling distribution ( x ) = 63.5
standard Deviation ( sd )= 2.9/ Sqrt ( 7 ) =1.0961
sample size (n) = 7
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 63) = (63-63.5)/2.9/ Sqrt ( 7 )
= -0.5/1.0961
= -0.4562
= P ( Z <-0.4562) From Standard Normal Table
= 0.3241
P(X < 64) = (64-63.5)/2.9/ Sqrt ( 7 )
= 0.5/1.0961 = 0.4562
= P ( Z <0.4562) From Standard Normal Table
= 0.6759
P(63 < X < 64) = 0.6759-0.3241 = 0.3517
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