Chapter 2.1, Problem 3E

(a)

To determine

Antawn’s percentile having head circumference of 22.4 inches.

Answer to Problem 3E

Percentile that corresponds with head circumference of 22.4 inches is 33^rd percentile.

Explanation of Solution

Given information:

Measurements of head circumference (in inches):

  

Let

n , Number of data values that are less than the individual’s data value

N , Total number of data values

Now,

Sort all the data values in ascending order:

20.8, 20.8, 21.0, 21.5, 21.6, 21.7, 21.9, 22.0, 22.2, 22.3, 22.4, 22.5, 22.5, 22.6, 22.6, 22.7, 22.7, 22.7, 23.0, 23.0, 23.1, 23.3, 23.4, 23.5, 23.5, 23.9, 23.9, 24.0, 24.2, 25.6.

Note that

10 of the 30 data values are smaller than 22.4.

Percentile can be determined as the number of data values that are less than the individual’s data value ( n ), divided by the total number of data values ( N ), written as percentage.

  $\begin{array}{l}\text{Percentile}=\frac{n}{N}\times 100\%\\ =\frac{10}{30}\times 100\%\\ =0.33\times 100\%\\ =33\%\end{array}$

Thus,

Percentile that corresponds with head circumference of 22.4 inches is 33^rd percentile.

(b)

To determine

Head circumference of the player at the 90^th percentile of the distribution.

Answer to Problem 3E

Head circumference at the 90^thpercentile is 24.0 inches.

Explanation of Solution

Given information:

Measurements of head circumference (in inches):

  

Let

n , Number of data values that are less than the individual’s data value

N , Total number of data values

Now,

Sort all the data values in ascending order:

20.8, 20.8, 21.0, 21.5, 21.6, 21.7, 21.9, 22.0, 22.2, 22.3, 22.4, 22.5, 22.5, 22.6, 22.6, 22.7, 22.7, 22.7, 23.0, 23.0, 23.1, 23.3, 23.4, 23.5, 23.5, 23.9, 23.9, 24.0, 24.2, 25.6.

Note that

The data set consists of 30 data values.

Such that

  $N=30$

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

Then

90^th percentile has 90% of the 30 data values below it.

Calculate 90% of 30 data values:

  $n=90\%\times N=\frac{90}{100}\times 30=\frac{9}{10}\times 30=27$

We have

27 out of 30 data values are smaller than 90^th percentile.

90^th percentile needs to be the 28^th data value in the ascending order, and that is 24.0.

Thus,

Head circumference of the player at the 90^thpercentile of the distribution is 24.0 inches.