Question

The lengths of adult males' hands are normally distributed with mean 186 mm and standard deviation is 7.1 mm. Suppose that 10 individuals are randomly chosen. Round all answers to 4 where possible.

1. What is the distribution of ¯xx¯? ¯xx¯ ~ N( , )
2. For the group of 10, find the probability that the average hand length is more than 185.
3. Find the third quartile for the average adult male hand length for this sample size.
4. For part b), is the assumption that the distribution is normal necessary?

Answer 1

Solution:

Given :
n=10,
mean=mu=186 mm
Population standard deviation=sigma=7.1mm
A) For tha sample of size 10,the sample mean is same as the population mean
muX_bar = mu=186mm and
Sampling standard deviation will be ,
SigmaX_ bar=sigma/✓n
=7.1/✓10
=2.2452
Xbar~ N(muX_bar=186,sigmaX_bar=2.2452)
B) P(Xbar>185)=1-P(Xbar<=185)
=1- p{[(Xbar - muX_bar)/sigmaX_bar] -[(185-186)/2.2452]}
=1-p( z<=−0.4454)
=1-0.328
=0.672
C) To find the 3rd qurtile ,we have to find the corresponding

z-values as
P( Z>=z)= 75%
p( Z>=z)= 0.75
P( Z>= 0.6745)= 0.75
From the standard normal table.
Now using the z-score formula, we can find the value of Xbar as follows,
z=(Xbar-muX_bar)/sigmaX_bar
z× SigmaX_ bar=(Xbar-muX_bar)
Xbar= muX_bar+( z×SigmaX_ bar)
x= 186+(0.6745×2.2452)
=186+(1.5144)
x=187.5144
The 3rd quartile is 187.5144
D) yes, the assumption that the distribution is normal necessary so that Xbar also have normal distribution.