Question

gebra i b-cr
pre-test complete
time remain
55.02
grace is looking at a report of her monthly cell-phone usage for the last year to determine if she needs to upgrade her plan. the list represents the approximate number of megabytes of data grace used each month.
700, 735, 680, 890, 755, 740, 670, 785, 805, 1050, 820, 750
what is the standard deviation of the data? round to the nearest whole number.
○ 65
○ 75
○ 100
○ 130

Explanation:

Step1: Calculate the mean

First, sum all data points, then divide by the count (12 months).
Sum = $700 + 735 + 680 + 890 + 755 + 740 + 670 + 785 + 805 + 1050 + 820 + 750 = 9380$
Mean $\bar{x} = \frac{9380}{12} \approx 781.67$

Step2: Find squared differences

Subtract mean from each data point, square the result.
$(700-781.67)^2 \approx 6669.99$, $(735-781.67)^2 \approx 2178.09$, $(680-781.67)^2 \approx 10336.79$,
$(890-781.67)^2 \approx 11735.39$, $(755-781.67)^2 \approx 711.29$, $(740-781.67)^2 \approx 1736.39$,
$(670-781.67)^2 \approx 12470.19$, $(785-781.67)^2 \approx 11.09$, $(805-781.67)^2 \approx 544.29$,
$(1050-781.67)^2 \approx 71999.09$, $(820-781.67)^2 \approx 1469.19$, $(750-781.67)^2 \approx 1002.99$

Step3: Calculate variance

Sum squared differences, divide by data count.
Sum of squares $\approx 6669.99+2178.09+10336.79+11735.39+711.29+1736.39+12470.19+11.09+544.29+71999.09+1469.19+1002.99 = 120864.28$
Variance $\sigma^2 = \frac{120864.28}{12} \approx 10072.02$

Step4: Compute standard deviation

Take square root of variance, round to whole number.
$\sigma = \sqrt{10072.02} \approx 100.36 \approx 100$

Answer:

100 (Option C: 100)