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Calculate the mean and standard deviation for the following table giventhe age distribution of a group of people:Age:20-3030-4040-5050-6060-7070-8080-90No. of persons:351122141130512

Answer» First we have to construct a table for the given data. Please refer to video for creating complete table.
From the table,
`sum f_i u_i = 25`
`sum f_i u_i^2 = 705`
`N = sum f_i = 500``:.` Variance `(sigma^2)= [(sum f_i u_i^2)/N - ((sum f_i u_i)/N)^2]h^2`
Here, `h = ` Class size `= 10`
`:. sigma^2= [705/500 - (25/500)^2]*10^2 = 10^2(1.4075)`
`:.` Standard deviation `(sigma) = sqrt(10^2(1.4075)) = 10**1.186 = 11.86`
Now, `Mean(barX) = A+h[(sum f_i u_i)/N]`
Here, `A = ` Assumed Mean ` = 55`
`:. barX = 55+ 10(25/500) = 55.5`
`:.` Mean is `55.5`.


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