Question

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

Answer 1

Answer:

See explanation below.

Step-by-step 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:

$s^2 = \frac{\sum f X_m^2 -[\frac{(\sum f Xm)^2}{n}]}{n-1}$

And if we replace we got:

$s^2 = \frac{2203400 -\frac{(13020)^2}{98}}{97}= 4882.474$

And then the standard deviation is given bY:

$s = \sqrt{4882.474}=69.87$

So then the best answer would be : 70