Answer to Question #120179 in Statistics and Probability for Mary Abena Asabea

Question

The shelf life of a particular diary product is is normal distributed with the mean of 12 days and a variance of 9 days. About what percentage of the product will last between 13 to 16 days?

Answer 1

We will find a probability for a product to last between 13 and 16 days. This corresponds to the respective percentage. Let $X$ be variable that corresponds to the shelf life of the product. It is normally distributed with $\mu=12,\,\, \sigma=9$ . Its probability density function is $p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}$ . Then

$P(13\leq X\leq16)=\int_{13}^{16}\frac{1}{9\sqrt{2\pi}}e^{-\frac12(\frac{x-12}{9})^2}dx\approx0.1274$

The latter is  rounded to 4 decimal places. This corresponds to 12,74% of product.

The latter integral was computed with the help of Anaconda (free distribution of

the Python). The following code was used in Jupyter Notebook (a part of the distribution

for writing and running code) for calculation of the integral:

_{from scipy import integrate}

_{import numpy as np}

_import math

_{func = lambda x:(1/(9*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-12)/9)*((x-12)/9))}

_{Pr = integrate.quad(func, 13, 16)}

_print(Pr)

Answer:12,74%