Question

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, 850, 755, 740, 670, 785, 805, 1050, 820, 750. what is the standard deviation of the data? round to the nearest whole number. 65 75 100 120

Explanation:

Step1: Calculate the mean

First, find the sum of the data values: $700 + 735+680 + 890+755+740+670+785+805+1050+820+750=9280$. There are $n = 12$ data - points. The mean $\bar{x}=\frac{9280}{12}\approx773.33$.

Step2: Calculate the squared differences

For each data - point $x_i$, calculate $(x_i-\bar{x})^2$. For example, for $x_1 = 700$, $(700 - 773.33)^2=(-73.33)^2 = 5377.2989$. Do this for all 12 data - points and sum them up. The sum of squared differences $\sum_{i = 1}^{12}(x_i-\bar{x})^2=107966.67$.

Step3: Calculate the variance

The variance $s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{107966.67}{11}\approx9815.15$.

Step4: Calculate the standard deviation

The standard deviation $s=\sqrt{s^2}=\sqrt{9815.15}\approx99.07\approx100$.

Answer:

100