Question

Batter up! Refer to Exercise $48$Use the $68–95–99.7$rule to answer the following questions.

a. About what percent of Major League Baseball players with $100$ plate appearances had batting averages of $0.363$or higher? Show your method clearly.

b. A player with a batting average of $0.227$ is at about what percentile in this distribution? Justify your answer.

Short answer

Part (a) $0.15\%$ of the Major League Basketball players with $100$ plate appearances had batting averages of$0.363$ and higher.

Part (b) The player with a batting average of $0.227$ is at $16th$ percentile.

Step-by-step solution

Part (a) Step 1: Given information

Mean,$\mu =0.261$

Standard deviation,$\sigma =0.034$

Part (a) Step 2: Concept

Graph of normal probability- We can claim that the distribution is roughly Normal if the Normal probability plot has a linear structure.

Part (a) Step 3: Calculation

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

$0.363$lies $3\sigma$above the mean.

$\mu +3\sigma =0.261+3(0.034)=0.363$

According to $68-95-99.7$rule:

$99.7\%$of the data values lie within $3\sigma$of the mean.

Although,

Data values in total are $100\%$

Then

$100\%-99.7\%=0.30\%$

$0.30\%$of the data values lie more than $3\sigma$from the mean.

We also know that

Around the mean, the normal distribution is symmetric.

That implies

$0.15\%$of the data values are more than $3\sigma$below the mean.

And

$0.15\%$of the data values are more than $3\sigma$above the mean.

Therefore,

With $100$plate appearances, $0.15$percent of Major League Basketball players had batting averages of $0.363$or better.

Part (b) Step 1: Calculation

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

$0.227$lies $\sigma$($1$standard deviation) below the mean.

$\mu -\sigma =0.261-0.034=0.227$

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

Around the mean, the normal distribution is symmetric.

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 $Xth$percentile includes $x\%$of the data values below it.

That implies

$16\%$of the players have a batting average of $0.227$or less.

Thus,

The player with a batting average of $0.227$is at $16th$percentile.