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Finite Math Examples
ClassFrequency90-99480-89670-79460-69350-59240-491ClassFrequency90−99480−89670−79460−69350−59240−491
Step 1
Reorder the classes with their related frequencies in an ascending order (lowest number to highest), which is the most common.
ClassFrequency(f)40-49150-59260-69370-79480-89690-994ClassFrequency(f)40−49150−59260−69370−79480−89690−994
Step 2
Step 2.1
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)LowerLimitsUpperLimits40-491404950-592505960-693606970-794707980-896808990-9949099ClassFrequency(f)LowerLimitsUpperLimits40−491404950−592505960−693606970−794707980−896808990−9949099
Step 2.2
The class midpoint is the lower class limit plus the upper class limit divided by 22.
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40-491404940+49250-592505950+59260-693606960+69270-794707970+79280-896808980+89290-994909990+992ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40−491404940+49250−592505950+59260−693606960+69270−794707970+79280−896808980+89290−994909990+992
Step 2.3
Simplify all the midpoint column.
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40-491404944.550-592505954.560-693606964.570-794707974.580-896808984.590-994909994.5ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40−491404944.550−592505954.560−693606964.570−794707974.580−896808984.590−994909994.5
Step 2.4
Add the midpoints column to the original table.
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5ClassFrequency(f)Midpoint(M)40−49144.550−59254.560−69364.570−79474.580−89684.590−99494.5
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5ClassFrequency(f)Midpoint(M)40−49144.550−59254.560−69364.570−79474.580−89684.590−99494.5
Step 3
Calculate the square of each group midpoint M2M2.
ClassFrequency(f)Midpoint(M)M240-49144.544.5250-59254.554.5260-69364.564.5270-79474.574.5280-89684.584.5290-99494.594.52ClassFrequency(f)Midpoint(M)M240−49144.544.5250−59254.554.5260−69364.564.5270−79474.574.5280−89684.584.5290−99494.594.52
Step 4
Simplify the M2M2 column.
ClassFrequency(f)Midpoint(M)M240-49144.51980.2550-59254.52970.2560-69364.54160.2570-79474.55550.2580-89684.57140.2590-99494.58930.25ClassFrequency(f)Midpoint(M)M240−49144.51980.2550−59254.52970.2560−69364.54160.2570−79474.55550.2580−89684.57140.2590−99494.58930.25
Step 5
Multiply each midpoint squared by its frequency ff.
ClassFrequency(f)Midpoint(M)M2f⋅M240-49144.51980.251⋅1980.2550-59254.52970.252⋅2970.2560-69364.54160.253⋅4160.2570-79474.55550.254⋅5550.2580-89684.57140.256⋅7140.2590-99494.58930.254⋅8930.25ClassFrequency(f)Midpoint(M)M2f⋅M240−49144.51980.251⋅1980.2550−59254.52970.252⋅2970.2560−69364.54160.253⋅4160.2570−79474.55550.254⋅5550.2580−89684.57140.256⋅7140.2590−99494.58930.254⋅8930.25
Step 6
Simplify the f⋅M2 column.
ClassFrequency(f)Midpoint(M)M2f⋅M240-49144.51980.251980.2550-59254.52970.255940.560-69364.54160.2512480.7570-79474.55550.252220180-89684.57140.2542841.590-99494.58930.2535721
Step 7
Find the sum of all frequencies. In this case, the sum of all frequencies is n=1,2,3,4,6,4=20.
∑f=n=20
Step 8
Find the sum of f⋅M2 column. In this case, 1980.25+5940.5+12480.75+22201+42841.5+35721=121165.
∑f⋅M2=121165
Step 9
Step 9.1
Reorder the classes with their related frequencies (ƒ) in an ascending order (lowest number to highest), which is the most common.
ClassFrequency(f)40-49150-59260-69370-79480-89690-994
Step 9.2
Find the midpoint M for each class.
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5
Step 9.3
Multiply the frequency of each class by the class midpoint.
ClassFrequency(f)Midpoint(M)f⋅M40-49144.51⋅44.550-59254.52⋅54.560-69364.53⋅64.570-79474.54⋅74.580-89684.56⋅84.590-99494.54⋅94.5
Step 9.4
Simplify the f⋅M column.
ClassFrequency(f)Midpoint(M)f⋅M40-49144.544.550-59254.510960-69364.5193.570-79474.529880-89684.550790-99494.5378
Step 9.5
Add the values in the f⋅M column.
44.5+109+193.5+298+507+378=1530
Step 9.6
Add the values in the frequency column.
n=1+2+3+4+6+4=20
Step 9.7
The mean (mu) is the sum of f⋅M divided by n, which is the sum of frequencies.
μ=∑f⋅M∑f
Step 9.8
The mean is the sum of the product of the midpoints and frequencies divided by the total of frequencies.
μ=153020
Step 9.9
Simplify the right side of μ=153020.
76.5
76.5
Step 10
The equation for the standard deviation is S2=∑f⋅M2-n(μ)2n-1.
S2=∑f⋅M2-n(μ)2n-1
Step 11
Substitute the calculated values into S2=∑f⋅M2-n(μ)2n-1.
S2=121165-20(76.5)220-1
Step 12
Simplify the right side of S2=121165-20(76.5)220-1 to get the variance S2=216.84210526.
216.84210526