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 4242 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6 comma 8046,804 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 804 l b over 42 endfraction equals 162 6,804 lb 42=162 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 174174 lb and a standard deviation of 36.936.9.

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.

Step-by-step explanation:

X the number of correct answers the dog chooses by guessing

Then each guess is independent of the other.  Also there are only two outcomes either correct or not correct.

Since dog does not know the answer and there is simple guessing the dog selecting the correct answer = dog correcting noncorrect answer = 1/2

X is binomial (n =10 and p = 1/2)

Expected values of correct answers

=E(X)

=Mean of a binomial variable

=np

Here n =10 and p =1/2

Hence expected value of correct answers = 5