With the frequecy distribution shown in the 50 cities of pakistan,
- range = 18
- mean = 18.1
- median = 19.8333
- mode = 19.125
- kurtosis = 2.7508
- Standard deviation = 3.75
How to find the Range
= highest value - lowest value
= 26.5 - 8.5
= 18
How to find the mean
= ∑ f x / ∑ f
= ∑ f x / N
= 905 / 50
= 18.1
median
= lower limit + ( N/2 - C ) * h / ( frequency of the class interval )
C = cumulative frequency preceeding to the median class frequency
h = class interval
= 18.5 + ( 50 / 2 - ( 5 + 12 ) ) * 3 / 18
= 18.5 + 1.3333
= 19.8333
How to find the mode
The mode is the value with the highest frequency occurence. This is under class 18 - 20
mode = lower limit + ( ( f1 - f0 ) / (2*f1 - f0 - f2 ) ) * h
f1 = fequency of the modal class
f0 = freqency of the preceeding modal class
f2 = frequency of the next modal class
h = class interval
= 18.5 + ( ( 18 - 12 ) / (2 * 18 - 12 - 8 ) ) * 3
= 18 + ( 0.375 ) * 3
= 19.125
How to find the standard deviation
= sqrt ( 1 / N ∑ f ( x - x' )^2 )
= sqrt (1 / 50 * 706.5
= 3.7589
How to solve for relative dispersion
= standard deviation / mean
= 3.7589 / 3
= 1.2530
What is the variance?
= ( standard deviation )^2
= ( 3.7589 )^2
= 14.1293
How to solve for kurtosis
= ∑ f ( x - x' )^4 / ( N * ( standard deviation )^4 )
= 27459.405 / ( 50 * 3.7589^4 )
= 2.7509
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