Question

An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have only one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags.a) The average baggage-related revenue per passenger is: $(please round to the nearest cent)b) The standard deviation of baggage-related revenue is: $(please round to the nearest cent)c) About how much revenue should the airline expect for a flight of 120 passengers?(please round to the nearest dollar)

Answer 1

Answer:

(a) The average baggage-related revenue per passenger is $15.70.

(b) The standard deviation of baggage-related revenue is $19.95.

(c) The expected revenue for 120 passengers is $1,884.

Step-by-step explanation:

Let the random variable X represent the baggage-related revenue per passenger.

It is provided that the baggage fees is $25 for the first bag and $35 for the second.

The probability model is:

    X :   0         25      60

P (X) : 0.54    0.34    0.12

(a)

Compute the average baggage-related revenue per passenger as follows:

$E(X)=\sum {X\times P (X)}$

         $=(0\times 0.54)+(25\times 0.34)+(60\times 0.12)\\\\=0+8.50+7.20\\\\=15.70$

Thus, the average baggage-related revenue per passenger is $15.70.

(b)

Compute the standard deviation of baggage-related revenue as follows:

$E (X^{2})=\sum X^{2}\times P(X)}\\\\=(0^{2}\times 0.54)+(25^{2}\times 0.34)+(60^{2}\times 0.12)\\\\=0+212.50+432\\\\=644.5$

$SD(X)=\sqrt{E(X^{2})-(E(X))^{2}}$

           $=\sqrt{644.5-(15.7)^{2}}\\\\=19.950188\\\\\approx 19.95$

Thus, the standard deviation of baggage-related revenue is $19.95.

(c)

Compute the expected revenue for 120 passengers as follows:

$E(R)=n\times E(X)$

        $=120\times 15.7\\\\=1884$

Thus, the expected revenue for 120 passengers is $1,884.