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
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