Chapter 2.2, Problem 51E

(a)

To determine

Percentage of bags weigh less than 9.02 ounces.

Answer to Problem 51E

2.5% of the bags weigh less than 9.02 ounces.

Explanation of Solution

Given information:

Mean, $\mu =9.12$

Standard deviation, $\sigma =0.05$

According to 68 − 95 − 99.7 rule:

68% of the data of a normal distribution lies with 1 standard deviation from the mean.

95% of the data of a normal distribution lies with 2 standard deviation from the mean.

99.7% of the data of a normal distribution lies with 1 standard deviation from the mean.

Then

The general Normal density graph is represented as:

  

Note that

9.02 lies $2\sigma$ below the mean.

  $\mu -2\sigma =9.12-2\left(0.05\right)=9.02$

According to 68 − 95 − 99.7 rule:

95% of the data values lie within $2\sigma$ of the mean.

Although,

Data values in total are 100%.

Then

  $100\%-95\%=5\%$

5% of the data values lie more than $2\sigma$ from the mean.

We also know that

Normal distribution is symmetric about the mean.

That implies

2.5% of the data values are more than $2\sigma$ above the mean.

And

2.5% of the data values are more than $2\sigma$ below the mean.

Therefore,

2.5% of the bags weigh less than 9.02 ounces.

(b)

To determine

Percentile of the bag in the distribution that weighs 9.07 ounces.

Answer to Problem 51E

The bag is at 16^th percentile.

Explanation of Solution

Given information:

Mean, $\mu =9.12$

Standard deviation, $\sigma =0.05$

According to 68 − 95 − 99.7 rule:

68% of the data of a normal distribution lies with 1 standard deviation from the mean.

95% of the data of a normal distribution lies with 2 standard deviation from the mean.

99.7% of the data of a normal distribution lies with 1 standard deviation from the mean.

Then

The general Normal density graph is represented as:

  

Note that

9.02 lies $\sigma$ below the mean.

  $\mu -\sigma =9.12-0.05=9.07$

According to 68 − 95 − 99.7 rule:

68% of the data values lie within $\sigma$ (1 standard deviation) of the mean.

Although,

Data values in total are 100%.

Then

  $100\%-68\%=32\%$

32% of the data values lie more than $\sigma$ (1 standard deviation) from the mean.

We also know that

Normal distribution is symmetric about the mean.

That implies

16% of the data values are more than $\sigma$ (1 standard deviation) above the mean.

And

16% of the data values are more than $\sigma$ (1 standard deviation) below the mean.

The data value represented by the x^th percentile includes x% of the data values below it.

That implies

Bag that weighs 9.07 ounces has about 16% of the other weighs below it.

Thus,

The bag is at 16^th percentile.