Question

christine is currently taking a college astronomy class and the instructor often gives quizzes. on the past seven quizzes, christine got the scores shown below. find the standard deviation, rounding to one more decimal place than is present in the original data.
50 15 31 27 11 42 71
(1 point)
20.9
437.6
31
8715.6

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here \(n = 7\), \(x_{1}=50\), \(x_{2}=15\), \(x_{3}=31\), \(x_{4}=27\), \(x_{5}=11\), \(x_{6}=42\), \(x_{7}=71\).
\(\sum_{i=1}^{7}x_{i}=50 + 15+31+27+11+42+71=247\)
\(\bar{x}=\frac{247}{7}\approx35.3\)

Step2: Calculate the variance

The formula for the variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\((x_{1}-\bar{x})^{2}=(50 - 35.3)^{2}=14.7^{2}=216.09\)
\((x_{2}-\bar{x})^{2}=(15 - 35.3)^{2}=(- 20.3)^{2}=412.09\)
\((x_{3}-\bar{x})^{2}=(31 - 35.3)^{2}=(-4.3)^{2}=18.49\)
\((x_{4}-\bar{x})^{2}=(27 - 35.3)^{2}=(-8.3)^{2}=68.89\)
\((x_{5}-\bar{x})^{2}=(11 - 35.3)^{2}=(-24.3)^{2}=590.49\)
\((x_{6}-\bar{x})^{2}=(42 - 35.3)^{2}=6.7^{2}=44.89\)
\((x_{7}-\bar{x})^{2}=(71 - 35.3)^{2}=35.7^{2}=1274.49\)
\(\sum_{i = 1}^{7}(x_{i}-\bar{x})^{2}=216.09+412.09+18.49+68.89+590.49+44.89+1274.49 = 2625.53\)
\(s^{2}=\frac{2625.53}{7 - 1}=\frac{2625.53}{6}\approx437.6\)

Step3: Calculate the standard deviation

The formula for the standard deviation \(s=\sqrt{s^{2}}\)
\(s=\sqrt{437.6}\approx20.9\)

Answer:

20.9