Question

Before every​ flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. the aircraft can carry 4040 ​passengers, and a flight has fuel and baggage that allows for a total passenger load of 6 comma 7606,760 lb. the pilot sees that the plane is full and all passengers are men. the aircraft will be overloaded if the mean weight of the passengers is greater than startfraction 6 comma 760 l b over 40 endfraction equals 169 6,760 lb 40=169 lb. what is the probability that the aircraft is​ overloaded? should the pilot take any action to correct for an overloaded​ aircraft? assume that weights of men are normally distributed with a mean of 180.2180.2 lb and a standard deviation of 3838.

Answer 1

Answer:

Step-by-step explanation:

X, the mean weight of passengers is N(180.2, 3838)

No of passengers n = 4040

std error = STd dev/ sqrt n

= 60.38

The probability that the aircraft is overloaded is 0.9920; since this is a greater than 50% chance, the pilot should either reduce the number of passengers on the plane or require them to reduce their baggage.

The z-score for this is given by

z = (X-μ)/σ = (169-180.2)/60.38 = -11.2/60.38 = -0.02

Using a z-table

P(Z<-0.02) = 0.0080

Prob for aircraft to be overloaded

= P(Z>-0.02)

=1-0.0080

=0.9920

Answer 2

The probability that the aircraft is overloaded is 0.6141; since this is a greater than 50% chance, the pilot should either reduce the number of passengers on the plane or require them to reduce their baggage.

The z-score for this is given by

z = (X-μ)/σ = (169-180.2)/38 = -11.2/38 = -0.29.

Using a z-table (http://www.z-table.com) we see that the probability of weights being less than this (the area to the left of this score) is 0.3859; to find the probability that the weight is greater than this (to the right), we subtract from 1:

1-0.3859 = 0.6141.