IB Mathematics SL 4.8 Binomial distribution. Mean and variance-AI HL Paper 2- Exam Style Questions- New Syllabus

Question

The aircraft for a particular flight has 72 seats. The airline’s records show that historically for this flight only 90% of the people who purchase a ticket arrive to board the flight. They assume this trend will continue and decide to sell extra tickets and hope that no more than 72 passengers will arrive. The number of passengers that arrive to board this flight is assumed to follow a binomial distribution with a probability of 0.9.

(a) The airline sells 74 tickets for this flight. Find the probability that more than 72 passengers arrive to board the flight.

(b)
(i) Write down the expected number of passengers who will arrive to board the flight if 72 tickets are sold.
(ii) Find the maximum number of tickets that could be sold if the expected number of passengers who arrive to board the flight must be less than or equal to 72.

(c) Each passenger pays \$150 for a ticket. If too many passengers arrive, then the airline will give \$300 in compensation to each passenger that cannot board. Find, to the nearest integer, the expected increase or decrease in the money made by the airline if they decide to sell 74 tickets rather than 72.

▶️ Answer/Explanation

Markscheme

(a)
    Model: $ X \sim \text{Bin}(74, 0.9) $, where $ X $ is the number of passengers who arrive
    We need $ P(X > 72) = P(X = 73) + P(X = 74) $
    Using binomial probability formula:
    $ P(X = 73) = \binom{74}{73} (0.9)^{73} (0.1)^1 = 74 \times (0.9)^{73} \times 0.1 \approx 0.00338014 $
    $ P(X = 74) = \binom{74}{74} (0.9)^{74} (0.1)^0 = (0.9)^{74} \approx 0.000411098 $
    $ P(X > 72) \approx 0.00338014 + 0.000411098 \approx 0.00379124 $
    Alternatively, using GDC: $ P(X > 72) = 1 – P(X \leq 72) \approx 0.00379 $ (A1)
Result:
0.00379

(b)(i)
For $ X \sim \text{Bin}(72, 0.9) $:
Expected value: $ E[X] = n \times p = 72 \times 0.9 = 64.8 $ (A1)
Result:
64.8 passengers

(b)(ii)
    Require: $ E[X] = n \times 0.9 \leq 72 $
    $ n \leq \frac{72}{0.9} = 80 $
    Since $ n $ must be an integer, the maximum is: $ n = 80 $ (A1)
Result:
80 tickets

(c)
    Method 1:
    Either:
    When selling 74 tickets

	$ T \leq 72 $	$ T = 73 $	$ T = 74 $
Income minus compensation (I)	11100	10800	10500
Probability	$ 0.9962\ldots $	$ 0.003380\ldots $	$ 0.0004110\ldots $

(A1)(A1)
    $ E(I) = 11100 \times 0.9962\ldots + 10800 \times 0.00338\ldots + 10500 \times 0.000411 \approx 11099 $ (A1)
    Income for 72 tickets = $ 72 \times 150 = 10800 $ (A1)
    Expected gain = $ 11099 – 10800 = 299 $ (A1)
    Or:
    Income is $ 74 \times 150 = 11100 $ (A1)
    Expected compensation is:
    $ 0.003380\ldots \times 300 + 0.0004110\ldots \times 600 = 1.26070\ldots $ (M1)(A1)(A1)
    Expected income when selling 74 tickets is $ 11100 – 1.26070\ldots = 11098.73\ldots \approx 11099 $ (M1)(A1)
    Then:
    Income for 72 tickets = $ 72 \times 150 = 10800 $ (A1)
    So expected gain $ \approx 11099 – 10800 = 299 $ (A1)
    Method 2:
    For 74 tickets sold, let $ C $ be the compensation paid out
    $ P(T = 73) = 0.00338014\ldots $, $ P(T = 74) = 0.000411098\ldots $
    $ E(C) = 0.003380\ldots \times 300 + 0.0004110\ldots \times 600 = 1.26070\ldots $ (M1)(A1)(A1)
    Extra expected revenue = $ 300 – 1.01404\ldots – 0.246658\ldots = 299 $ (A1)
    Method 3:
    Let $ D $ be the change in income when selling 74 tickets

	$ T \leq 72 $	$ T = 73 $	$ T = 74 $
Change in income	300	0	-300

(A1)(A1)
$ P(T \leq 73) = 0.9962\ldots $, $ P(T = 74) = 0.000411098\ldots $
$ E(D) = 300 \times 0.9962\ldots + 0 \times 0.003380\ldots – 300 \times 0.0004110 = 299 $ (M1)(A1)(A1)
Result:
\$299 increase

	\( T \leq 72 \)	\( T = 73 \)	\( T = 74 \)
Income minus compensation (I)	11100	10800	10500
Probability	\( 0.9962\ldots \)	\( 0.003380\ldots \)	\( 0.0004110\ldots \)

IB Mathematics SL 4.8 Binomial distribution. Mean and variance-AI HL Paper 2- Exam Style Questions- New Syllabus

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