Question

(2c) The Shelf life of a particular dairy product is normally distributed with a mean of 12 days and a
variance of 9 days. About what percentage of the products last between 13 and 16 days.

Answer 1

It is given that Shelf life is Normally Distributed with mean 12 and variance 9

Let Shelf life be denoted by X

then

$X \sim N(\mu = 12, \sigma^2 = 9)$

And we need to find the probability such that

$P(13<X<16) = P(\frac{13 - \mu}{\sigma}< \frac{X-\mu}{\sigma} < \frac{16 - \mu}{\sigma}) = P(\frac{13 - 12}{3}< Z < \frac{16 - 12}{3})$ =

$= P(1/3<Z<4/3) = P(Z<4/3) - P(Z<1/3) = \phi(4/3) - \phi(1/3) =$  0.9087888 - 0.6305587 =  0.2782301

where $\phi(.)$ is the CDF of standard normal distribution and these values are calculated from standard normal table.