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

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=(X-\mu)/(\sigma)`

When X=9 it follows:

`\begin{gathered} z=(9-12)/(3) \\ z=(-3)/(3) \\ z=-1 \end{gathered}`

When X=15, it follows:

`\begin{gathered} z=(15-12)/(3) \\ z=(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<p></p><p>b. the probability that the product last between 12 and 15 days is given as:</p>[tex]P(12The z value for each value of x is[tex]\begin{gathered} x=12\Rightarrow z=(12-12)/(3)=0 \\ x=15\Rightarrow z=(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<p></p><p>c. the probability that the product last less than 6 days is given as:</p>[tex]P(X<6)`

The z value is

`\begin{gathered} z=(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=(15-12)/(3) \\ z=(3)/(3) \\ z=1 \end{gathered}`

Therefore the probability becomes:

`P(z>1)`

The probability is

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