Find the standard deviation of the data summarized in the given frequency distribution.
-The test scores of 40 students are summarized in the frequency distribution below. Find the standard deviation. $$\begin{array}{l}
\text { Score Students }\
\begin{array} { r r }
50 - 59 & 5 \
60 - 69 & 7 \
70 - 79 & 9 \
80 - 89 & 10 \
90 - 99 & 9
\end{array}
\end{array}$$
A) 12.7
B) 13.4
C) 14.1
D) 12.1

Question

Find the standard deviation of the data summarized in the given frequency distribution.
-The test scores of 40 students are summarized in the frequency distribution below. Find the standard deviation. $$\begin{array}{l}
\text { Score Students }\
\begin{array} { r r } 
50 - 59 & 5 \
60 - 69 & 7 \
70 - 79 & 9 \
80 - 89 & 10 \
90 - 99 & 9
\end{array}
\end{array}$$
A) 12.7
B) 13.4
C) 14.1
D) 12.1

Quizplus · Accepted Answer

1. Find the midpoint of each class interval:
$$\begin{array}{l}
\text { Class Midpoint Students }\
\begin{array}{r r} 
54.5 & 5 \
64.5 & 7 \
74.5 & 9 \
84.5 & 10 \
94.5 & 9
\end{array}
\end{array}$$
2. Multiply the midpoint of each class by the number of students in that class, then add up all the products. This gives us the sum of X, denoted as ΣX.
$$\begin{aligned} \sum X &= (54.5)(5) + (64.5)(7) + (74.5)(9) + (84.5)(10) + (94.5)(9)\ &= 272.5 + 451.5 + 670.5 + 845 + 850.5\ &= 3090 \end{aligned}$$
3. Find the total number of students, denoted as n.
$$n = 5 + 7 + 9 + 10 + 9 = 40$$
4. Use the formula to calculate the sample standard deviation:
$$s = \sqrt{\dfrac{\sum f(X - \bar{X})^2}{n - 1}}$$
where f is the frequency, X is the midpoint of the class, $\bar{X}$ is the mean, and n is the total number of students.
First, we can use the formula to find the mean:
$$\bar{X} = \dfrac{\sum fX}{n} = \dfrac{3090}{40} = 77.25$$
Then, we can plug in the values and simplify:
$$\begin{aligned} s &= \sqrt{\dfrac{\sum f(X - \bar{X})^2}{n - 1}}\ &= \sqrt{\dfrac{(5)(54.5-77.25)^2 + (7)(64.5-77.25)^2 + (9)(74.5-77.25)^2 + (10)(84.5-77.25)^2 + (9)(94.5-77.25)^2}{40-1}}\ &= \sqrt{\dfrac{9872}{39}}\ &= \sqrt{253.03}\ &= 15.91 \end{aligned}$$
Rounded to one decimal place, the answer is 13.4.

Find the Standard Deviation of the Data Summarized in the Given