Statistics Examples
ClassFrequency12-17318-23624-29430-352ClassFrequency12−17318−23624−29430−352
Step 1
Step 1.1
The lower limit for every class is the smallest value in that class. On the other hand, the upper limit for every class is the greatest value in that class.
ClassFrequency(f)LowerLimitsUpperLimits12-173121718-236182324-294242930-3523035ClassFrequency(f)LowerLimitsUpperLimits12−173121718−236182324−294242930−3523035
Step 1.2
The class midpoint is the lower class limit plus the upper class limit divided by 22.
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)12-173121712+17218-236182318+23224-294242924+29230-352303530+352ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)12−173121712+17218−236182318+23224−294242924+29230−352303530+352
Step 1.3
Simplify all the midpoint column.
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)12-173121714.518-236182320.524-294242926.530-352303532.5ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)12−173121714.518−236182320.524−294242926.530−352303532.5
Step 1.4
Add the midpoints column to the original table.
ClassFrequency(f)Midpoint(M)12-17314.518-23620.524-29426.530-35232.5ClassFrequency(f)Midpoint(M)12−17314.518−23620.524−29426.530−35232.5
ClassFrequency(f)Midpoint(M)12-17314.518-23620.524-29426.530-35232.5ClassFrequency(f)Midpoint(M)12−17314.518−23620.524−29426.530−35232.5
Step 2
Calculate the square of each group midpoint M2M2.
ClassFrequency(f)Midpoint(M)M212-17314.514.5218-23620.520.5224-29426.526.5230-35232.532.52ClassFrequency(f)Midpoint(M)M212−17314.514.5218−23620.520.5224−29426.526.5230−35232.532.52
Step 3
Simplify the M2M2 column.
ClassFrequency(f)Midpoint(M)M212-17314.5210.2518-23620.5420.2524-29426.5702.2530-35232.51056.25ClassFrequency(f)Midpoint(M)M212−17314.5210.2518−23620.5420.2524−29426.5702.2530−35232.51056.25
Step 4
Multiply each midpoint squared by its frequency ff.
ClassFrequency(f)Midpoint(M)M2f⋅M212-17314.5210.253⋅210.2518-23620.5420.256⋅420.2524-29426.5702.254⋅702.2530-35232.51056.252⋅1056.25ClassFrequency(f)Midpoint(M)M2f⋅M212−17314.5210.253⋅210.2518−23620.5420.256⋅420.2524−29426.5702.254⋅702.2530−35232.51056.252⋅1056.25
Step 5
Simplify the f⋅M2f⋅M2 column.
ClassFrequency(f)Midpoint(M)M2f⋅M212-17314.5210.25630.7518-23620.5420.252521.524-29426.5702.25280930-35232.51056.252112.5ClassFrequency(f)Midpoint(M)M2f⋅M212−17314.5210.25630.7518−23620.5420.252521.524−29426.5702.25280930−35232.51056.252112.5
Step 6
Find the sum of all frequencies. In this case, the sum of all frequencies is n=3,6,4,2=15n=3,6,4,2=15.
∑f=n=15∑f=n=15
Step 7
Find the sum of f⋅M2f⋅M2 column. In this case, 630.75+2521.5+2809+2112.5=8073.75630.75+2521.5+2809+2112.5=8073.75.
∑f⋅M2=8073.75∑f⋅M2=8073.75
Step 8
Step 8.1
Find the midpoint MM for each class.
ClassFrequency(f)Midpoint(M)12-17314.518-23620.524-29426.530-35232.5ClassFrequency(f)Midpoint(M)12−17314.518−23620.524−29426.530−35232.5
Step 8.2
Multiply the frequency of each class by the class midpoint.
ClassFrequency(f)Midpoint(M)f⋅M12-17314.53⋅14.518-23620.56⋅20.524-29426.54⋅26.530-35232.52⋅32.5ClassFrequency(f)Midpoint(M)f⋅M12−17314.53⋅14.518−23620.56⋅20.524−29426.54⋅26.530−35232.52⋅32.5
Step 8.3
Simplify the f⋅Mf⋅M column.
ClassFrequency(f)Midpoint(M)f⋅M12-17314.543.518-23620.512324-29426.510630-35232.565ClassFrequency(f)Midpoint(M)f⋅M12−17314.543.518−23620.512324−29426.510630−35232.565
Step 8.4
Add the values in the f⋅Mf⋅M column.
43.5+123+106+65=337.543.5+123+106+65=337.5
Step 8.5
Add the values in the frequency column.
n=3+6+4+2=15n=3+6+4+2=15
Step 8.6
The mean (mu) is the sum of f⋅Mf⋅M divided by nn, which is the sum of frequencies.
μ=∑f⋅M∑fμ=∑f⋅M∑f
Step 8.7
The mean is the sum of the product of the midpoints and frequencies divided by the total of frequencies.
μ=337.515μ=337.515
Step 8.8
Simplify the right side of μ=337.515μ=337.515.
22.522.5
22.522.5
Step 9
The equation for the standard deviation is S2=∑f⋅M2-n(μ)2n-1S2=∑f⋅M2−n(μ)2n−1.
S2=∑f⋅M2-n(μ)2n-1S2=∑f⋅M2−n(μ)2n−1
Step 10
Substitute the calculated values into S2=∑f⋅M2-n(μ)2n-1S2=∑f⋅M2−n(μ)2n−1.
S2=8073.75-15(22.5)215-1S2=8073.75−15(22.5)215−1
Step 11
Simplify the right side of S2=8073.75-15(22.5)215-1S2=8073.75−15(22.5)215−1 to get the variance S2=34.‾285714S2=34.¯¯¯¯¯¯¯¯¯¯¯¯285714.
34.2857142834.28571428
Step 12
The standard deviation is the square root of the variance 34.‾28571434.¯¯¯¯¯¯¯¯¯¯¯¯285714. In this case, the standard deviation is 5.855400435.85540043.
5.855400435.85540043