we have that
For 95% confidence interval
Using a Z-scores table
the value of z=1.96
Remember that
z =(x - μ)/σ
we have
μ=5.8 in
σ=0.9 in
substitute
For z=1.96
1.96=(x-5.8)/0.9
x=1.96*0.9+5.8
x=7.564 in
For z=-1.96
-1.96=(x-5.8)/0.9
x=-1.96*0.9+5.8
x=4.036 in
therefore
between 4.036 in and 7.564 in
Part 2
If you were to draw samples of size 42 from this population, in what range would you expect to find the middle 95% of most averages for the breadths of male heads in the sample?
In this part, divide the standard deviation by the square root of the sample size
![\frac{0.9}{\sqrt[\square]{42}}=0.1389](https://img.qammunity.org/2023/formulas/mathematics/college/8rkfgedat8g06r6zv4f9e9kqsv3n23o7is.png)
the middle is equal to the average of
5.8(+/-)1.96*0.1389
-----> 6.07
-----> 5.53
Find out the average
(6.07+5.5278)/2=5.8
the middle is equal to the mean in this problem
between 5.57 and 6.07