Question

Suppose you actually had 200 independent units in the sample and calculated the sample mean of 91.15g. The sample standard deviation is 0.9 grams. How many units from your new sample of 200 do you expect to lie in your 68% tolerance interval? Your 95% tolerance interval? Your 99.7% tolerance interval? Do not round any of your answers. Hint: Lecture 15 example 2 part d) should help you here.

Answer 1

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 91.15
standard Deviation ( sd )= 0.9

a.
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [91.15 ± 0.9]
= [ 91.15 - 0.9 , 91.15 + 0.9]
= [ 90.25 , 92.05 ]
b.
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [91.15 ± 2 * 0.9]
= [ 91.15 - 2 * 0.9 , 91.15 + 2* 0.9]
= [ 89.35 , 92.95 ]
c.
About 99.7% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [91.15 ± 3 * 0.9]
= [ 91.15 - 3 * 0.9 , 91.15 + 3* 0.9]
= [ 88.45 , 93.85 ]