The population of potential clientele have head breadths normally distributed with mean 6.9 in and standard deviation 0.8 in
98% of the breadths must differ from the mean by no more than 2.33 times the standard deviation.
This mean the the range we would expect to find the middle 98% of most head breadths is represented by the interval [6.9 - 2.33*0.8, 6.9 + 2.33*0.8] = [5.04, 8.76] in
If we are considering samples of size 47, the standard deviation for the mean of each sample is given by:
![\frac{\sigma}{\sqrt[]{n}}=\frac{0.8}{\sqrt[]{47}}=0.117\text{ in}](https://img.qammunity.org/2023/formulas/mathematics/college/dli3f79n2p81098u2gv6216du71bgqasdh.png)
In this case, we would expect to find 98% of most averages for the breadths of male heads in the interval [6.9 - 2.33*0.117, 6.9 + 2.33*0.117] = [6.63, 7.17] in