# given the following frequency distribution find the mean, variance and standard deviation Class            31-33    34-36   37-39   40-42   43-45 Frequency       18        18        10       15         5 Take the mean value of each class i to represent the value of that class. In this case, this is the middle value of each class.

The mean is given by

`Sigma_(i=1)^5 (x_i f(x_i))/N`

where `x_i` is the class value of the `i`th class , `f(x_i)` is the frequency of...

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Take the mean value of each class i to represent the value of that class. In this case, this is the middle value of each class.

The mean is given by

`Sigma_(i=1)^5 (x_i f(x_i))/N`

where `x_i` is the class value of the `i`th class , `f(x_i)` is the frequency of that class value and N is the total frequency over all the 5 classes.

`therefore`  mean `bar x` = (32(18) + 35(18) + 38(10) + 41(15) + 44(5))/(18 + 18 + 10 + 15 + 5)

= 2421/66 = 36.7

The variance is given by

`Sigma_(i=1)^5 (f(x_i)(x_i - barx)^2)/N`

`= (18(32-36.7)^2 + 18(35-36.7)^2 + 10(38-36.7)^2 + 15(41-36.7)^2 + 5(44-36.7)^2)/66`

`= 1010.34/66 = 15.3`

The standard deviation is the square root of the variance,

so the standard deviation = `sqrt(15.3) = 3.91`

mean = 36.7, variance = 15.3 and standard deviation = 3.91

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