Chapter 12.3, Problem 18PPS

a.

To determine

To calculate : the mean and standard deviation of the given data.

Answer to Problem 18PPS

The mean height of the players is $\underset{\_}{79\text{ inches}}$ .

The standard deviation of the given data is $\underset{\_}{4.1}$ .

Explanation of Solution

Given information :

Consider the information given in the question:

The heights (in inches) of fifteen players in a NBA team:

  $80,80,78,77,83,85,83,74,81,74,83,81,78,69,79$

Formula used :

Step1: Find the mean, $\overline{x}$

Step2: Find the square of the difference between each value in the set of data and the mean. Then sum the squares, and divide by the number of values in the set of data. The result is the variance.

Step3: Take the square root of the variance to find the standard deviation.

Calculation :

As per problem,

The data set: $80,80,78,77,83,85,83,74,81,74,83,81,78,69,79$

The number of values in the data set $\text{=15}$

Step1: To find the mean, add the heights (in inches) and then divide by the number of values in the data set.

  $\begin{array}{l}\overline{x}=\frac{80+80+78+77+83+85+83+74+81+74+83+81+78+69+79}{15}\\ \overline{x}=\frac{1185}{15}=79\end{array}$

Therefore, the mean height of fifteen NBA players is $79\text{ inches}$ .

Step2: To find the variance, square the difference between each value and the mean. Then add the squares, and divide by the number of values.

  $\begin{array}{}$ {\sigma }^2=[{(80-79)}^2+{(80-79\text{)}}^2\text{+(}78-79{\text{)}}^2\text{+(77}-79{\text{)}}^2\text{+(83}-79{\text{)}}^2\text{+}$$ \text{ (85}-79{\text{)}}^2\text{+(83}-79{\text{)}}^2\text{+(}74-79{\text{)}}^2\text{+(81}-79{\text{)}}^2\text{+(}74-79{\text{)}}^2+$$ {(83-79)}^2+{(81-79)}^2+{(78-79)}^2+{(69-79)}^2+{(79-79)}^2]÷15$$ =[{(1)}^2+{(1)}^2+{(-1)}^2+{(-2)}^2+{(4)}^2+{(6)}^2+{(4)}^2+{(-5)}^2+{(2)}^2$$ {(-5)}^2+{(4)}^2+{(2)}^2+{(-1)}^2+{(-10)}^2+{(0)}^2]÷15$$ =\frac{1+1+1+4+16+36+16+25+4+25+16+4+1+100+0}{15}$$ =\frac{250}{15}=16.67$\end{array}$

Step 3: the standard deviation is the square root of the variance

  $\begin{array}{l}{\sigma }^2=16.67\\ \sqrt{{\sigma }^2}=\sqrt{16.67}\\ \sigma \approx 4.1\end{array}$

b.

To determine

To calculate: the mean and standard deviation of the data after trade of players and describe the change in calculations.

Answer to Problem 18PPS

The mean height of the players is $\underset{\_}{78.73\text{ inches}}$ .

The standard deviation of the given data is $\underset{\_}{4.8}$ .

Due to trade of players the mean height decreases and the standard deviation increases.

Explanation of Solution

Given:

The heights (in inches) of $\text{15}$ players in a NBA team are:

  $80,80,78,77,83,85,83,74,81,74,83,81,78,69,79$

A player of $5\text{'}9"$ is traded for Earl Boykins.

The height of Earl Boykins is $65\text{ inches}$ .

Formula used :

Step1: Find the mean, $\overline{x}$

Step2: Find the square of the difference between each value in the set of data and the mean. Then sum the squares, and divide by the number of values in the set of data. The result is the variance.

Step3: Take the square root of the variance to find the standard deviation.

Calculation:

As per the given problem,

The height of the player who was traded $=5\text{'}9"=69\text{ inches}$

Replace the height of the traded player from the given data with the height of Earl Boykins.

The new set of data is:

  $80,80,78,77,83,85,83,74,81,74,83,81,78,\underset{\_}{65},79$

The number of values in the data set $\text{=15}$

Step1: To find the mean, add the heights (in inches) and then divide by the number of values in the data set.

  $\begin{array}{l}\overline{x}=\frac{80+80+78+77+83+85+83+74+81+74+83+81+78+65+79}{15}\\ \overline{x}=\frac{1181}{15}\approx 78.73\end{array}$

Therefore, the new mean height of fifteen NBA players is about $78.73\text{ inches}$ .

Step2: To find the new variance, square the difference between each value and the mean. Then add the squares, and divide by the number of values.

  $\begin{array}{}$ {\sigma }^2=[{(80-78.73)}^2+{(80-78.73\text{)}}^2\text{+(}78-78.73{\text{)}}^2\text{+(77}-78.73{\text{)}}^2\text{+(83}-78.73{\text{)}}^2\text{+}$$ \text{ (85}-78.73{\text{)}}^2\text{+(83}-78.73{\text{)}}^2\text{+(}74-78.73{\text{)}}^2\text{+(81}-78.73{\text{)}}^2\text{+(}74-78.73{\text{)}}^2+$$ {(83-78.73)}^2+{(81-78.73)}^2+{(78-78.73)}^2+{(65-78.73)}^2+{(79-78.73)}^2]÷15$$ =[{(1.27)}^2+{(1.27)}^2+{(-0.73)}^2+{(-1.73)}^2+{(4.27)}^2+{(6.27)}^2+{(4.27)}^2+{(-4.73)}^2$$ +{(2.27)}^2\text{+ }{(-4.73)}^2+{(4.27)}^2+{(2.27)}^2+{(-0.73)}^2+{(-13.73)}^2+{(0.27)}^2]÷15$$ =[1.61+1.61+0.53+2.99+18.23+39.31+18.23+22.37+5.15+22.37+$$ 18.23+5.15+0.53+188.51+0.07]÷15$$ =\frac{344.89}{15}\approx 22.99$\end{array}$

Step 3: the new standard deviation is the square root of the variance

  $\begin{array}{l}{\sigma }^2=22.99\\ \sqrt{{\sigma }^2}=\sqrt{22.99}\\ \sigma \approx 4.8\end{array}$

The meanand standard deviation of the given data before the trade was $79\text{ inches}$ and $4.1$ (as calculated in part (a)).

The mean and standard deviation of the given data after the trade is $78.73\text{ inches}$ and $4.8$ respectively.

Therefore, the mean height decreases due to trade of players,but the standard deviation increases.

c.

To determine

To convert:each data value into centimeters and find the mean.

To convert the calculated mean to inches.

To compare the calculated mean (in inches) to that mean calculated in part (a).

Answer to Problem 18PPS

The mean height of the players is $\underset{\_}{200.66\text{ cm}}$ .

The mean height in inches is $\underset{\_}{79\text{ inches}}$ , which is same as the mean calculated in part (a).

Explanation of Solution

Given:

The heights (in inches) of $\text{15}$ players in a NBA team are:

  $80,80,78,77,83,85,83,74,81,74,83,81,78,69,79$

Formula used :

To find the mean, $\overline{x}$ ,add the heights (in centimeters) and then divide by the number of values in the data set.

Calculation:

As per the given problem,

The heights (in inches) of $\text{15}$ players in a NBA team are:

  $80,80,78,77,83,85,83,74,81,74,83,81,78,69,79$

To convert the height of each player into centimeter, use conversion relation: $1\text{ inch}=2.54\text{ cm}$

Thereforethe height in centimeters:

  $203.2,203.2,198.12,195.58,210.82,215.9,210.82,187.96,205.74,187.96,210.82,205.74,198.12,175.26,200.66$

To calculate the mean, add the heights (in centimeters) and then divide by the number of values in the data set.

  $\begin{array}{l}\overline{x}= [203.2+203.2+198.12+195.58+210.82+215.9+210.82+187.96+\\ 205.74+187.96+210.82+205.74+198.12+175.26+200.66]÷15\\ \overline{x}=\frac{3009.9}{15}\approx 200.66\end{array}$

Therefore, the mean height (in centimeters) of fifteen NBA players is about $200.66\text{ cm}$ .

To convert the mean (in centimeters) to inches, use conversion relation $1\text{ cm}=\frac{1}{2.54}\text{ inches}$

Therefore, $\overline{x}=200.66\text{ cm}=\frac{200.66}{2.54}\text{ inches}=79\text{ inches}$

The calculated mean in inches is same as the mean calculated in part (a).