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 
3039 
4049 
5059 
6069 
7079 
8089 
9099 


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.
x  f  f*M  f*M^{2} 
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 s^{2}=(ΣfM^{2}ΣfM^{2}/n)/(n1)=  185.1489  
Std deviation s=  √s^{2} =  13.6069 
from above
Standard deviation =13.6
option B is correct
B> The computed value is not significantly different from the given value
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