Consider the following test scores obtained by a class: $\text { cellpadding="0" cellspacing }=" 0 " \text { width }=52 \% \text { " border }=0 "$
$$\begin{array} { l l }
\text { Test Score } & \text { Frequency } \
90 & 2 \
80 & 5 \
70 & 10 \
60 & 5 \
50 & 3
\end{array}$$ Find the variance and standard deviation, rounded to the nearest hundredth.
A) variance: $119.36$ standard deviation: $10.93$
B) variance: $10.93$ standard deviation: $119.36$
C) variance: $69.20$ standard deviation: $10.93$
D) variance: $69.20$ standard deviation: $119.36$
E) variance: $100.00$ standard deviation: $10.00$

Question

Consider the following test scores obtained by a class: $\text { cellpadding="0" cellspacing }=" 0 " \text { width }=52 \% \text { " border }=0 ">$
$$\begin{array} { l l } 
\text { Test Score } & \text { Frequency } \
90 & 2 \
80 & 5 \
70 & 10 \
60 & 5 \
50 & 3
\end{array}$$ Find the variance and standard deviation, rounded to the nearest hundredth.
A) variance: $119.36$ standard deviation: $10.93$
B) variance: $10.93$ standard deviation: $119.36$
C) variance: $69.20$ standard deviation: $10.93$
D) variance: $69.20$ standard deviation: $119.36$
E) variance: $100.00$ standard deviation: $10.00$

Quizplus · Accepted Answer

To find the variance and standard deviation, we need to first find the mean:
$$\bar{x}=\frac{(90 \cdot 2)+(80 \cdot 5)+(70 \cdot 10)+(60 \cdot 5)+(50 \cdot 3)}{25}=70$$
Now, we can use the formulas:
$$\text {Variance}=\frac{\sum(x_i-\bar{x})^2}{n}=\frac{(90-70)^2\cdot2+(80-70)^2\cdot5+(70-70)^2\cdot10+(60-70)^2\cdot5+(50-70)^2\cdot3}{25}=119.36$$
$$\text{Standard deviation}=\sqrt{\text{Variance}}=\sqrt{119.36}=10.93$$
Therefore, the answer is A.

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