Question

use the display of data to find the standard deviation.
the standard deviation is approximately □.
(do not round until the final answer. then round to the nearest hundredth as needed.)

Explanation:

Step1: Calculate the mid - point of each class and the total frequency

Let \(x_i\) be the mid - point of each class and \(f_i\) be the frequency.
For class \(7\): \(x_1 = 7\), \(f_1=11\)
For class \(8\): \(x_2 = 8\), \(f_2 = 8\)
For class \(9\): \(x_3=9\), \(f_3 = 11\)
The total frequency \(n=\sum_{i = 1}^{3}f_i=11 + 8+11=30\)

Step2: Calculate the mean \(\bar{x}\)

The mean \(\bar{x}=\frac{\sum_{i = 1}^{3}f_ix_i}{n}\)
\(\sum_{i = 1}^{3}f_ix_i=(7\times11)+(8\times8)+(9\times11)=77 + 64+99 = 240\)
\(\bar{x}=\frac{240}{30}=8\)

Step3: Calculate \(\sum_{i = 1}^{3}f_i(x_i-\bar{x})^2\)

For \(i = 1\): \(f_1(x_1 - \bar{x})^2=11\times(7 - 8)^2=11\times1 = 11\)
For \(i = 2\): \(f_2(x_2-\bar{x})^2=8\times(8 - 8)^2=8\times0 = 0\)
For \(i = 3\): \(f_3(x_3-\bar{x})^2=11\times(9 - 8)^2=11\times1=11\)
\(\sum_{i = 1}^{3}f_i(x_i-\bar{x})^2=11 + 0+11=22\)

Step4: Calculate the standard deviation \(s\)

The formula for the sample standard deviation (when data is from a sample, and since we are not told it's a population, we use the sample formula \(s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}\)
\(s=\sqrt{\frac{22}{30 - 1}}=\sqrt{\frac{22}{29}}\approx\sqrt{0.7586}\approx0.87\)

Answer:

\(0.87\)