Question

Find the standard​ deviation, s, of sample data summarized in the frequency distribution table given 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,

9.09.0.

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	2020​-2626	2727​-3333	3434​-4040	4141​-4747	4848​-5454	5555​-6161
Frequency	55	1616	4242	1616	88	33

Standard

deviationequals=nothing

​(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,

9.09.0​?

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

C.

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

Answer 1

mid-point	frequency
x	f	f*M	f*M2
23	5	115	2645
30	16	480	14400
37	42	1554	57498
44	16	704	30976
51	8	408	20808
58	3	174	10092
total	90	3435	136419
mean =x̅=Σf*M/Σf=	38.1667
sample Var s2=(ΣfM2-ΣfM2/n)/(n-1)=			59.7360
Std deviation s=		√s2 =	7.7289

from above:

Standard deviation s =7.7

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