Problem Statement
The masses of apples, $X$ grams, from a commercial orchard are normally distributed with mean $\mu = 180\text{ g}$ and standard deviation $\sigma = 15\text{ g}$.
(a) Find $P(X < 160)$.
(b) The orchard classifies apples into small, medium, and large. The heaviest $15\%$ of apples are classified as large. Find the minimum mass of a large apple.
(c) A random sample of 8 apples is selected. Find the probability that exactly 2 apples are classified as large.
Verified Solution & Marking Scheme
Evaluate P(X < 160) using GDC normalCDF
Using GDC normalCDF(lower: $-10^{99}$, upper: 160, $\mu: 180, \sigma: 15$):
$$Z = \frac{160 - 180}{15} = -\frac{20}{15} = -1.333$$
$$P(X < 160) = 0.091211... \approx 0.0912\text{ (3 s.f.)}$$
Find threshold k for heaviest 15% using invNorm
If the heaviest $15\%$ are large, then $P(X > k) = 0.15 \implies P(X \le k) = 0.85$.
Using GDC invNorm(area: $0.85, \mu: 180, \sigma: 15$):
$$k = 180 + 1.03643(15) = 180 + 15.546 = 195.546... \approx 196\text{ g (3 s.f.)}$$
Model sample using Binomial distribution B(8, 0.15)
Let $Y$ be the number of large apples in the sample of 8.
$Y \sim B(n = 8, p = 0.15)$.
$$P(Y = 2) = \binom{8}{2} (0.15)^2 (0.85)^6 = 28 \times 0.0225 \times 0.3771495 = 0.2376... \approx 0.238\text{ (3 s.f.)}$$