9.5000032 is one such number
Answer: (751.05, 766.95)
Step-by-step explanation:
We know that the confidence interval for population mean is given by :-
,
where
=population standard deviation.
= sample mean
n= sample size
z* = Two-tailed critical z-value.
Given : 
n= 42

We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576
Now, the 99% confidence interval around the true population mean viscosity :-
![759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)](https://tex.z-dn.net/?f=759%5Cpm%20%282.5760%29%5Cdfrac%7B20%7D%7B%5Csqrt%7B42%7D%7D%5C%5C%5C%5C%3D759%5Cpm%20%282.5760%29%283.086067%29%5C%5C%5C%5C%3D759%5Cpm7.9497%3D%28759-7.9497%2C%5C%20759%2B7.9497%29%5C%5D%5C%5C%3D%28751.0503%2C%5C%20766.9497%29%5Capprox%28751.05%2C%5C%20766.95%29)
∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)
LMP would be 74 degrees and NMP would be 63
Answer: 40
Step-by-step explanation: