A fisherman has 200 fish to sell with the length of the fishes given in the table below. What is the standard deviation of the lengths of the fish?Length Lcm,Frequency 0≤L<10     30...

A fisherman has 200 fish to sell with the length of the fishes given in the table below. What is the standard deviation of the lengths of the fish?

Length Lcm,Frequency

0≤L<10     30

10≤L<20    40

20≤L<30    50

30≤L<40    30

40≤L<60    33

60≤L<75    11

75≤L<100    6

Asked on by nicoleuc

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The lengths of the fish are provided as a table with the length divided into ranges and the number of fish that fall into each category.

To find the standard deviation, the steps to be followed have been illustrated with one range of the table. Consider the range [0-10], find the average length, here it is A1 = (0+10)/2 = 5. Multiply this with the number of fish which gives P1 = A1*N1 = 5*30 = 150. Also find the product S1 = A1^2*N1 = 25*30 = 750.

Length,Average,Frequency,Average*Frequency,Average^2*Frequency

[0 10]__5______30_____150____________750

[10 20)_15_____40_____600____________9000

[20 30)_25_____50_____1250___________31250

[30 40)_35_____30_____1050___________36750

[40 60)_50_____33_____1650___________82500

[60 75)_67.5____11____742.5___________50118.75

[75 100)_87.5___6_____525_____________45937.5

Add the corresponding average* frequency and divide the sum by the total number of fish. This gives the average: 5967.5/200 = 29.8375

Add the corresponding average^2*frequency and divide by the range: 256306.3/100 = 2563.063

The standard deviation is: `sqrt(2563.063 - 29.8375^2)`

=> 40.89971

The required standard deviation is 40.89971

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