Problem Statement
The weights of organic apples harvested in an orchard are normally distributed with mean $\mu = 180\text{ g}$ and unknown standard deviation $\sigma$. It is known that $8\%$ of the apples weigh more than $210\text{ g}$.
**(a)** [3 marks] Show that the standard deviation $\sigma$ is approximately $21.35\text{ g}$.
**(b)** [3 marks] Apples weighing less than $150\text{ g}$ are classified as small and sent for juicing. Calculate the probability that a randomly chosen apple is sent for juicing.
Verified Solution & Marking Scheme
Part (a) [M1]
Let $X \sim N(180, \sigma^2)$. $P(X > 210) = 0.08 \implies P(X \le 210) = 0.92$
Part (a) [M1]
Using inverse normal on GDC: $Z = \text{invNorm}(0.92, 0, 1) \approx 1.4051$
Part (a) [A1]
$Z = \frac{X - \mu}{\sigma} \implies 1.4051 = \frac{210 - 180}{\sigma} = \frac{30}{\sigma} \implies \sigma = \frac{30}{1.4051} \approx 21.35\text{ g}$ (AG)
Part (b) [M1]
Set up probability: $P(X < 150) = P\left(Z < \frac{150 - 180}{21.35}\right) = P(Z < -1.4051)$
Part (b) [M1]
Calculate using normalcdf on GDC: $\text{normalcdf}(-10^{99}, 150, 180, 21.35)$
Part (b) [A1]
$P(X < 150) = 0.0800$ (or by symmetry around mean, $0.08$)