Question

6. Is the standard deviation of the maximum drop distribution closer to 40, 70, or 100 hours? Explain your answer.

Answer 1

See explanation below.

Explanation:

Assuming the distribution on the figure attached. We have a distribution skewed to the right.

We can calculate the deviation with the following table:

Class frequency(f) (Xm) Midpoint f*Xm fXm^2

50-90 28 70 1960 137200

90-130 32 110 3520 387200

130-170 18 150 2700 405000

170-210 8 190 1520 288800

210-250 8 230 1840 423200

290-330 2 310 620 192200

410-450 2 430 860 369800

Total 98 13020 2203400

We are assuming the frequencies that;s important to mention.

We can calculate the sample variance with the following formula:

`s^2 = (\sum f X_m^2 -[((\sum f Xm)^2)/(n)])/(n-1)`

And if we replace we got:

`s^2 = (2203400 -((13020)^2)/(98))/(97)= 4882.474`

And then the standard deviation is given bY:

`s = √(4882.474)=69.87`

So then the best answer would be : 70