Chapter 3, Problem 3.2.21RE

a.

To determine

To calculate: The z-score for a temperature of $80°$.

Answer to Problem 3.2.21RE

The z-score for temperature $80°$ is 0.88.

Explanation of Solution

Given info:

The data represents the reported high temperatures of 23 cities of the United States in October.

Calculation:

Mean:

The sum of all the entries divided by the total number of entries is known as mean.

The mean is,

$\overline{X}=\frac{{\displaystyle \sum X}}{n}$

$\begin{array}{c}\overline{X}=\frac{\left(\begin{array}{l}62+72+66+79+83+61+62+85+\\ 72+64+74+71+42+38+91+66+\\ 77+90+74+63+64+68+42\end{array}\right)}{23}\\ =\frac{1,566}{23}\\ =68.0870\\ \approx 68.09\end{array}$

Thus, the mean for the data is 68.09.

The standard deviation is obtained using the table given below:

Data value	$X-\mu$	${\left(X-\mu \right)}^2$
62	$62-68.09=-6.09$	119.0281
72	$72-68.09=3.91$	222.3081
66	$66-68.09=-2.09$	50.2681
79	$79-68.09=10.91$	37.0881
83	$83-68.09=14.91$	285.9481
61	$61-68.09=-7.09$	15.2881
62	$62-68.09=-6.09$	16.7281
85	$85-68.09=16.91$	34.9281
72	$72-68.09=3.91$	8.4681
64	$64-68.09=-4.09$	680.6881
74	$74-68.09=5.91$	905.4081
71	$71-68.09=2.91$	524.8681
42	$42-68.09=-26.09$	4.3681
38	$38-68.09=-30.09$	79.3881
91	$91-68.09=22.91$	480.0481
66	$66-68.09=-2.09$	34.9281
77	$77-68.09=8.91$	25.9081
90	$90-68.09=21.91$	16.7281
74	$74-68.09=5.91$	0.0081
63	$63-68.09=-5.09$	680.6881
64	$64-68.09=-4.09$	119.0281
68	$68-68.09=-0.09$	222.3081
42	$42-68.09=-26.09$	50.2681
Total		4,223.082

Standard deviation:

The square root of the population variance is termed as population standard deviation.

$\begin{array}{c}\text{Population}\text{standard}\text{deviation}=\sigma \\ =\sqrt{{\sigma }^2}\\ =\sqrt{\frac{{\displaystyle \sum {\left(X-\mu \right)}^2}}{n}}\end{array}$

Substitute ${\displaystyle \sum {\left(X-\mu \right)}^2}$ as 4,223.082 and n as 23 in the formula,

$\begin{array}{c}\sigma =\sqrt{\frac{4,223.082}{23}}\\ =\sqrt{183.6123}\\ =13.5504\\ \approx 13.55\end{array}$

Thus the standard deviation of the data is 13.55.

Standard score:

The number of standard deviations that the value x lies from the mean μ represents the standard score or z-score.

$\begin{array}{c}z=\frac{\text{Value}-\text{Mean}}{\text{Standard deviation}}\\ =\frac{x-\mu }{\sigma }\end{array}$

• If the z-score falls outside the range –2 to 2 then it is considered as unusual.
• If the z-score falls outside the range –3 to 3 then it is considered as very unusual.

The z-score for 80ᵒ:

Here, mean is 68.09, standard deviation is 13.55 and x value as 80.

The z-score for $80°$ is,

$\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{80-68.09}{13.55}\\ =\frac{11.91}{13.55}\\ =0.8790\end{array}$

Thus, the z-score for temperature $80°$ is 0.88.

b.

To determine

To calculate: The z-score for temperature $56°$.

Answer to Problem 3.2.21RE

The z-score for temperature $56°$ is –0.89.

Explanation of Solution

Calculation:

From part (a) the mean and standard deviation is 68.09 and 13.55.

The z-score for temperature 56ᵒ:

Here, mean is 68.09, standard deviation is 13.55 and x value as 56.

The z-score for $56°$ is,

$\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{56-68.09}{13.55}\\ =\frac{-12.08}{13.55}\\ =-0.8915\end{array}$

Thus, the z-score for temperature 56ᵒ is –0.89.