Question

2. The shelf life of a particular dairy product is normally distributed with a mean of 12 days and a standarddeviation of 3 days.a) About what percent of the products last between 9 and 15 days?b) About what percent of the products last between 12 and 15 days?c) About what percent of the products last 6 days or less?d) About what percent of the products last 15 or more days?

Answer 1

SOLUTION

The given vlaues are:

$\mu=12,\sigma=3$

a. the probability that the product last between 9 and 15 days is given as:

$P(9Calculate the z value for each x value using[tex]z=\frac{X-\mu}{\sigma}$

When X=9 it follows:

$\begin{gathered} z=\frac{9-12}{3} \\ z=\frac{-3}{3} \\ z=-1 \end{gathered}$

When X=15, it follows:

$\begin{gathered} z=\frac{15-12}{3} \\ z=\frac{3}{3} \\ z=1 \end{gathered}$

Therefore the probability becomes

$P(-1Therefore the probabiolity is:[tex]P(-1Therefore about 68.3% of the products last between 9 and 15 days

b. the probability that the product last between 12 and 15 days is given as:

[tex]P(12The z value for each value of x is[tex]\begin{gathered} x=12\Rightarrow z=\frac{12-12}{3}=0 \\ x=15\Rightarrow z=\frac{15-12}{3}=1 \end{gathered}$

The probability becomes

$p(0Therefore the probability is:[tex]P(0Therefore about 34.1% of the products last between 12 and 15 days

c. the probability that the product last less than 6 days is given as:

[tex]P(X<6)$

The z value is

$\begin{gathered} z=\frac{6-12}{3} \\ z=-2 \end{gathered}$

Hence the probability is written as

$p(z<-2)$

Therefore about 2.31% of the products last 12 days or less

d. the probability that the product last 15 daysor more is given as

$P(X>15)$

The z value is:

$\begin{gathered} z=\frac{15-12}{3} \\ z=\frac{3}{3} \\ z=1 \end{gathered}$

Therefore the probability becomes:

$P(z>1)$

The probability is

$p(z>1)=15.1\%$