Chapter 14.3, Problem 8CFU

a.

To determine

To calculate: The mean, median, and standard deviation for the temperatures in Los Angeles.

Answer to Problem 8CFU

The mean , median , and standard deviation for the temperatures in Los Angeles are $70.375$ , $69.65$ , and $3.1352$ respectively.

Explanation of Solution

Given information:

The normal maximum daily temperature for Los Angeles.

Los Angeles	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	65.7	65.9	65.5	67.4	69.0	71.9	75.3	76.6	76.6	74.4	70.3	65.9

Formula used:

Formula of Mean

  $\overline{X}=\frac{{\displaystyle \sum X}}{n}$ where X represents each of the values in the data set.

Median for even terms = $\frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2}$

Median for odd terms = ${\left(\frac{n+1}{2}\right)}^{th}\text{term}$

A measure of variability associated with the arithmetic mean is the standard deviation.

  $\sigma =\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$

Calculation:

Consider the data in ascending order ,

  $\left\{65.5,65.7,65.9,65.9,67.4,69.0,70.3,71.9,74.4,75.3,76.6,76.6\right\}.$

So, $n=12$ .

The mean of the data is

  $\begin{array}{l}=\frac{65.5+65.7+65.9+65.9+67.4+69.0+70.3+71.9+74.4+75.3+76.6+76.6}{12}\\ =\frac{844.5}{12}\\ =70.375\end{array}$

There is even number of data.

The median of even number of data is −

  $\begin{array}{l}=\frac{{\left(\frac{12}{2}\right)}^{th}\text{term}+{\left(\frac{12}{2}+1\right)}^{th}\text{term}}{2}\\ =\frac{{\left(6\right)}^{th}\text{term}+{\left(7\right)}^{th}\text{term}}{2}\\ =\frac{69.0+70.3}{2}\\ =69.65\end{array}$

The standard deviation is −

$X_i$	$\overline{X}$	$X_i-\overline{X}$	${\left(X_i-\overline{X}\right)}^2$
65.5	70.375	$\begin{array}{l}=65.5-70.375\\ =-4.875\end{array}$	$\begin{array}{l}={\left(-4.875\right)}^2\\ =23.7656\end{array}$
65.7	70.375	$\begin{array}{l}=65.7-70.375\\ =-4.675\end{array}$	$\begin{array}{l}={\left(-4.675\right)}^2\\ =21.8556\end{array}$
65.9	70.375	$\begin{array}{l}=65.9-70.375\\ =-4.475\end{array}$	$\begin{array}{l}={\left(-4.475\right)}^2\\ =20.0256\end{array}$
65.9	70.375	$\begin{array}{l}=65.9-70.375\\ =-4.475\end{array}$	$\begin{array}{l}={\left(-4.475\right)}^2\\ =20.0256\end{array}$
67.4	70.375	$\begin{array}{l}=67.4-70.375\\ =-2.975\end{array}$	$\begin{array}{l}={\left(-2.975\right)}^2\\ =8.8506\end{array}$
69.0	70.375	$\begin{array}{l}=69.0-70.375\\ =-1.375\end{array}$	$\begin{array}{l}={\left(-1.375\right)}^2\\ =1.8906\end{array}$
70.3	70.375	$\begin{array}{l}=70.3-70.375\\ =-0.075\end{array}$	$\begin{array}{l}={\left(-0.075\right)}^2\\ =0.000005625\end{array}$
71.9	70.375	$\begin{array}{l}=71.9-70.375\\ =1.525\end{array}$	$\begin{array}{l}={\left(1.525\right)}^2\\ =2.3256\end{array}$
74.4	70.375	$\begin{array}{l}=74.4-70.375\\ =4.025\end{array}$	$\begin{array}{l}={\left(4.025\right)}^2\\ =16.2006\end{array}$
75.3	70.375	$\begin{array}{l}=75.3-70.375\\ =4.925\end{array}$	$\begin{array}{l}={\left(4.925\right)}^2\\ =24.2556\end{array}$
76.6	70.375	$\begin{array}{l}=76.6-70.375\\ =6.225\end{array}$	$\begin{array}{l}={\left(6.225\right)}^2\\ =38.7506\end{array}$
76.6	70.375	$\begin{array}{l}=76.6-70.375\\ =6.225\end{array}$	$\begin{array}{l}={\left(6.225\right)}^2\\ =38.7506\end{array}$
${\displaystyle \sum _{i=1}^{19}{\left(X_i-\overline{X}\right)}^2}$$$			$117.9574$

  $\begin{array}{l}=\sigma =\sqrt{\frac{1}{12}\left(117.9574\right)}\\ =\sigma =\sqrt{9.8297}\\ =\sigma =3.1352\end{array}$

Hence, the mean , median , and standard deviation for the temperatures in Los Angeles are $70.375$ , $69.65$ , and $3.1352$ respectively.

b.

To determine

To calculate: The mean, median, and standard deviation for the temperatures in Las Vegas.

Answer to Problem 8CFU

The mean , median , and standard deviation for the temperatures in Los Angeles are $80.48$ , $79.8$ , and $17.0551$ respectively.

Explanation of Solution

Given information:

The normal maximum daily temperature for Las Vegas .

Las Vegas	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	57.3	63.3	68.8	77.5	87.8	100.3	105.9	103.2	94.7	82.1	67.4	57.5

Formula used:

Formula of Mean

  $\overline{X}=\frac{{\displaystyle \sum X}}{n}$ where X represents each of the values in the data set.

Median for even terms = $\frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2}$

Median for odd terms = ${\left(\frac{n+1}{2}\right)}^{th}\text{term}$

A measure of variability associated with the arithmetic mean is the standard deviation.

  $\sigma =\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$

Calculation:

Consider the data in ascending order,

  $\left\{57.3,57.5,63.3,67.4,68.8,77.5,82.1,87.8,94.7,100.3,103.2,105.9\right\}.$

So, $n=12$ .

The mean of the data is

  $\begin{array}{l}=\frac{57.3+57.5+63.3+67.4+68.8+77.5+82.1+87.8+94.7+100.3+103.2+105.9}{12}\\ =\frac{965.8}{12}\\ =80.48\end{array}$

There is even number of data.

The median of even number of data is −

  $\begin{array}{l}=\frac{{\left(\frac{12}{2}\right)}^{th}\text{term}+{\left(\frac{12}{2}+1\right)}^{th}\text{term}}{2}\\ =\frac{{\left(6\right)}^{th}\text{term}+{\left(7\right)}^{th}\text{term}}{2}\\ =\frac{77.5+82.1}{2}\\ =79.8\end{array}$

The standard deviation is −

$X_i$	$\overline{X}$	$X_i-\overline{X}$	${\left(X_i-\overline{X}\right)}^2$
57.3	80.48	$\begin{array}{l}=57.3-80.48\\ =-23.18\end{array}$	$\begin{array}{l}={\left(-23.18\right)}^2\\ =537.3124\end{array}$
57.5	80.48	$\begin{array}{l}=57.5-70.375\\ =-22.98\end{array}$	$\begin{array}{l}={\left(-22.98\right)}^2\\ =528.0804\end{array}$
63.3	80.48	$\begin{array}{l}=63.3-80.48\\ =-17.18\end{array}$	$\begin{array}{l}={\left(-17.18\right)}^2\\ =295.1524\end{array}$
67.4	80.48	$\begin{array}{l}=67.4-80.48\\ =-13.08\end{array}$	$\begin{array}{l}={\left(-13.08\right)}^2\\ =171.0864\end{array}$
68.8	80.48	$\begin{array}{l}=68.8-80.48\\ =-11.68\end{array}$	$\begin{array}{l}={\left(-11.68\right)}^2\\ =136.4224\end{array}$
77.5	80.48	$\begin{array}{l}=77.5-80.48\\ =-2.98\end{array}$	$\begin{array}{l}={\left(-2.98\right)}^2\\ =8.8804\end{array}$
82.1	80.48	$\begin{array}{l}=82.1-80.48\\ =1.62\end{array}$	$\begin{array}{l}={\left(1.62\right)}^2\\ =2.6244\end{array}$
87.8	80.48	$\begin{array}{l}=87.8-80.48\\ =7.32\end{array}$	$\begin{array}{l}={\left(7.32\right)}^2\\ =53.5824\end{array}$
94.7	80.48	$\begin{array}{l}=94.7-80.48\\ =14.22\end{array}$	$\begin{array}{l}={\left(14.22\right)}^2\\ =202.2084\end{array}$
100.3	80.48	$\begin{array}{l}=100.3-80.48\\ =19.82\end{array}$	$\begin{array}{l}={\left(19.82\right)}^2\\ =392.8324\end{array}$
103.2	80.48	$\begin{array}{l}=103.2-80.48\\ =22.72\end{array}$	$\begin{array}{l}={\left(22.72\right)}^2\\ =516.1984\end{array}$
105.9	80.48	$\begin{array}{l}=105.9-80.48\\ =25.42\end{array}$	$\begin{array}{l}={\left(25.42\right)}^2\\ =646.1764\end{array}$
${\displaystyle \sum _{i=1}^{19}{\left(X_i-\overline{X}\right)}^2}$$$			$3490.5568$

Hence, the mean , median , and standard deviation for the temperatures in Los Angeles are $80.48$ , $79.8$ , and $17.0551$ respectively.

c.

To determine

To sketch : The box-and-whisker plot for the temperatures for each city.

Explanation of Solution

Given information:

The normal maximum daily temperature for Los Angeles and Las Vegas.

Los Angeles	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	65.7	65.9	65.5	67.4	69.0	71.9	75.3	76.6	76.6	74.4	70.3	65.9
Las Vegas	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	57.3	63.3	68.8	77.5	87.8	100.3	105.9	103.2	94.7	82.1	67.4	57.5

Graph:

  

Figure1.

  

Figure2.

Interpretation:

The figure1 and figure 2 represent the temperature of the Los Angeles and Las Vegas respectively.

d.

To determine

To show: Which of the city has a smaller variability in temperature.

Answer to Problem 8CFU

Las Vegas has a smaller variability in temperature.

Explanation of Solution

Given information:

The normal maximum daily temperature for Los Angeles and Las Vegas.

Los Angeles	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	65.7	65.9	65.5	67.4	69.0	71.9	75.3	76.6	76.6	74.4	70.3	65.9
Las Vegas	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	57.3	63.3	68.8	77.5	87.8	100.3	105.9	103.2	94.7	82.1	67.4	57.5

Calculation:

Record of Los Angeles

Terms	12
Mean	80.48
Median	79.8
Standard Deviation	17.0551

Record of Las Vegas

Terms	12
Mean	70.375
Median	69.65
Standard Deviation	3.1352

Observing the record , Las Vegas has a smaller variability in temperature. As each measure of Las Vegas is lesser than the Los Angeles.

e.

To determine

To show: The cause one city might have a greater variability in temperature than another.

Answer to Problem 8CFU

Pollution, temperature and many more.

Explanation of Solution

Given information:

The normal maximum daily temperature for Los Angeles and Las Vegas.

Los Angeles	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	65.7	65.9	65.5	67.4	69.0	71.9	75.3	76.6	76.6	74.4	70.3	65.9
Las Vegas	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
	57.3	63.3	68.8	77.5	87.8	100.3	105.9	103.2	94.7	82.1	67.4	57.5

Calculation:

Record of Los Angeles

Terms	12
Mean	80.48
Median	79.8
Standard Deviation	17.0551

Record of Las Vegas

Terms	12
Mean	70.375
Median	69.65
Standard Deviation	3.1352

The reason might the pollution, humidity, environment and could affect the temperature of the cities in large variation.