Question

Listed below are the top 10 annual salaries​ (in millions of​ dollars) of TV personalities. Find the​ range, variance, and standard deviation for the sample data. Given that these are the top 10​ salaries, do we know anything about the variation of salaries of TV personalities in​ general? 41 40 38 32 23 22 20 18 17.8 16.3 The range of the sample data is ​$ select: 24.7 24.7 million. ​(Type an integer or a​ decimal.) The variance of the sample data is 97.86. ​(Round to two decimal places as​ needed.) The standard deviation of the sample data is ​$ 9.89 million. ​(Round to two decimal places as​ needed.) Is the standard deviation of the sample a good estimate of the variation of salaries of TV personalities in​ general? A. ​No, because there is an outlier in the sample data. B. ​No, because the sample is not representative of the whole population. C. ​Yes, because the standard deviation is an unbiased estimator. D. ​Yes, because the sample is random.

Answer 1

$24.7million

$97.86million

$9.89million

Step-by-step explanation:

From the sample , the lowest number is 16.3 and the highest number is 41, the range is

41-16.3

=$24.7 million

Σ
`((x-x_(mean) )^2)/(n)`

In the sample given the mean is . : (41 +40 +38+ 32+ 23+ 22+ 20+ 18+ 17.8 +16.3 )/10

mean=26.81

Using that, we can find the variance:

[(41-26.81)^2+(40-26.81)^2+(38-26.81)^2+(32-26.81)^2+(23-26.81)^2+(22-26.81)^2+(20-26.81)^2+(18-26.81)^2+(17.8-26.81)^2+(16.3-26.81)^2]/10=97.86million

The standard deviation is just the square root of the variance:

standard deviation=√(var)

, the standard deviation is the square root of 97.86, which equals ​$ 9.89 million