Question

ind 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 -2929	3030 -3939	4040 -4949	5050 -5959	6060 -6969	7070 -7979
Frequency	44	2222	3636	2424	77	33

Standard deviationequals=

(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

Following table shows the calculations:

Interval
Lower	Upper	Mid Point, X	Frequency, f	fx	f(x-mean)^2
20	29	24.5	4	98	1895.7316
30	39	34.5	22	759	3047.7238
40	49	44.5	36	1602	112.7844
50	59	54.5	24	1308	1625.5896
60	69	64.5	7	451.5	2326.3303
70	79	74.5	3	223.5	2390.7987
Total			96	4442	11398.9584

So,

Now,

----------------------

A.

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