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	30-39	40​-49	50​-59	60​-69	70​-79	80​-89	90​-99
Frequency	1	1	7	4	9	32	34

1 Standard deviation

​(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 greater than the given value.The computed value is significantly greater than the given value.

B.

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

C.

The computed value is significantly less than the given value.

Answer 1

x	f	f*M	f*M2
34.5	1	34.5	1190.25
44.5	1	44.5	1980.25
54.5	7	381.5	20791.75
64.5	4	258	16641
74.5	9	670.5	49952.25
84.5	32	2704	228488
94.5	34	3213	303628.5
total	88	7306	622672
mean =x̅=Σf*M/Σf=	83.0227
sample Var s2=(ΣfM2-ΣfM2/n)/(n-1)=			185.1489
Std deviation s=		√s2 =	13.6069

from above

Standard deviation =13.6

option B is correct

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