2024 H2 Mathematics Paper 2 Question 9

The Binomial Distribution

Answers

Sketch.

$0.0125.$

The probability that each burger is found to have insufficient cheese in them is the same for each burger.
The event that a burger is found to have insufficient cheese in them is independent of all other burgers..

(ii)

$0.298.$

(iii)

$8.36 days .$

(iv)

$0.506.$

Full Solutions

(a)

binomial-distribution

(b)

Let $W$ be the random variable of the number of pies that are unsatisfactory out of $50$ and let $p$ be the probability that a randomly chosen pie is unsatisfactory

W \sim B (50, p)

By considering $E (W),$

\begin{matrix} n p \approx \frac{\sum f w}{\sum f} \\ (50) p \approx \frac{0 + (1) (9) + (2) (6) + (3) (4) + (4) (3) + (5) (1)}{80} \\ 50 p \approx \frac{5}{8} \\ p \approx 0.0125 ∎ \end{matrix}

(c)

(i)

The probability that each burger is found to have insufficient cheese in them is the same for each burger $∎$
The event that a burger is found to have insufficient cheese in them is independent of all other burgers. $∎$

(ii)

Y \sim B (120, 0.03)

\begin{aligned} P (Y < 3) \\ = P (Y \leq 2) \\ = 0.29844 \\ = 0.298 (3 s.f.) ∎ \end{aligned}

(iii)

Let $D$ denote the random variable of the number of days in which fewer than 3 burgers have insufficient cheese in them out of $28$

D \sim B (28, 0.29844)

\begin{aligned} E (D) \\ = n p \\ = (28) (0.29844) \\ = 8.3562 \\ = 8.36 days (3 s.f.) ∎ \end{aligned}

(iv)

Let $T$ denote the random variable of the number of burgers found to have insufficient cheese in them out of $120 \times 28 = 3360$ burgers in $28$ days

T \sim B (3360, 0.03)

\begin{aligned} P (T > 100) \\ = 1 - P (T \leq 100) \\ = 0.50578 \\ = 0.506 (3 s.f.) ∎ \end{aligned}