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

Answer 1

The answer is 20.9 when rounded it is 21

Answer 2

19.4

Explanation:

Given data,

50, 15, 31, 27, 11, 42, 71,

Let x represents the score,

Here, the number of scores, n = 7,

Thus, the mean score is,

`\bar{x}=(50 + 15 + 31 + 27 + 11 + 42 + 71)/(7)=(247)/(7)`

Hence, the standard deviation of the given data is,

`\sigma = \sqrt{\frac{\sum (x-\bar{x})^2}{n}}`

`=\sqrt{(\sum(x-(247)/(7))^2)/(7)}`

`=√(2625.4285714286){7}}`

`=√(375.0612244898)`

`=19.366497476\approx 19.4`