Stats Question--Need Help?

Question

Transportation authorities consider a particular 16-passenger airplane to be overloaded if the total carried exceeds 2,640 pounds. Suppose that the load placed on a plane by a typical passenger is normally distributed with a mean of 155 pounds and a standard deviation of 30 pounds. If passengers are typically assigned randomly to flights, without regard for their particular load, what proportion of flights will be overloaded?

Answer 1

Like that guy said before if the average person is more than 165 pounds the plane is overloaded.

Xbar is Normal ~ (the original mean, variance/n)

Xbar ~ N (155, 56.25)

P(Xbar > 165) = 165-155/(sqrt56.25) => z = 1.33

P(z > 1.33) = .0918

Answer 2

Weight Limit = 2640
16 passengers at mean weight of 155 = 2480

(2640 - 2480) = 160
So, if the total weight exceeds the average by 160 lbs (or 10 lbs per passenger), then the plane will be overloaded.

This means that if the average weight of passengers is 165 lbs (155 + 10), the plane will be overloaded.

Now, since you know the weights are normally distributed, what you need to do is find the total % of the population whose weight exceeds 165 lbs. You know the mean and standard deviation, which is enough information to fully specify the normal distribution function. Thre are two ways to do this:

1. Integrate the normal distribution from 165 to infinity using the mean of 155 and std. dev. of 30
2. Calculate (1 - integration of normal dist. from 0 to 165) using the mean of 155 and std. dev. of 30

Those will yield the same answer since the area under the normal curve is normalized to 1.

I plugged this into Excel (although you could integrate by hand or use a lookup table), and it yielded 36.94%

Answer 3

Let
Y = X_1 + X_2 + X_3 + ... + X_16
where X_i is the weight of passenger i. Since each X is normally distributed then the sum is normally distributed. We need to find the mean and std. deviation. I assume that each X is independent of the others.

mean(Y) = mean(X_1) + mean(X_2) + ... + mean(X_16)
= 155 + 155 + ... + 155
= 2480

Var(Y) = var(X_1) + var(X_2) + ... + var(X_16)
= 30^2 + 30^2 + ... + 30^2
= 16(30^2)
So the std. dev. of Y is
sqrt(16(30)^2) = 120.

Y is normally distributed with a mean of 2480 and a std. dev. of 120.
p(Y>2640) = p(Z>(2640-2480)/120)
=p(Z>1.3333)
=0.0912

Roughly 9.1% of the flights will be overloaded. I don't think I would want to fly on that airline!