Question

Find the standard​ deviation, s, of sample data summarized in the frequency distribution table below by using the formula​ below, where x represents the class​ midpoint, f represents the class​ frequency, and n represents the total number of sample values.​ Also, compare the computed standard deviation to the standard deviation obtained from the original list of data​ values,11.1

sequals=StartRoot StartFraction n left bracket Summation from nothing to nothing left parenthesis f times x squared right parenthesis right bracket minus left bracket Summation from nothing to nothing left parenthesis f times x right parenthesis right bracket squared Over n left parenthesis n minus 1 right parenthesis

EndFraction EndRootn∑f•x2−∑(f•x)2n(n−1)

Interval	20-26	27-33	34​-40	41​-47	48-54	55​-61	62​-68
Frequency	1	1	7	1	12	35	29

Standard deviations =____

​(Round to one decimal place as​ needed.)Consider a difference of​ 20% between two values of a standard deviation to be significant. How does this computed value compare with the given standard​ deviation,11.1​?

A. The computed value is significantly less than the given value.

B. The computed value is not significantly different from the given value.

C. The computed value is significantly greater than the given value.

Answer 1

The grouped data information provided, with the corresponding frequencies is shown in the table below:

Classes	Frequency (f)
20-26	1
27-33	1
34-40	7
41-47	1
48-54	12
55-61	35
62-68	29

For each of the grouped classes we need to compute the midpoint (M), by calculating the average between the lower and upper value of the class. Then, the following table si obtained:

Classes	M	f	M · f	M2 · f
20-26	23	1	23	529
27-33	30	1	30	900
34-40	37	7	259	9583
41-47	44	1	44	1936
48-54	51	12	612	31212
55-61	58	35	2030	117740
62-68	65	29	1885	122525
	Sum =	86	4883	284425

Calculated Standard deviation = 9.186

Given Standard deviation = 11.1

% difference between the two = 20.84%

A. The computed value is significantly less than the given value.

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