IBDP Maths SL 4.8 Binomial distribution, its mean and variance AA HL Paper 2- Exam Style Questions- New Syllabus

(a)
Let \( W \sim N(175, 8^2) \).
Using technology to find \( P(W < 170) \):
\( P(W < 170) \approx 0.265985 \).
Answer: \( \boxed{0.266} \) (26.6%)

(b)
Given \( P(W > w) = 0.2 \), which implies \( P(W < w) = 0.8 \).
Using the inverse normal function:
\( w \approx 181.732 \).
Answer: \( \boxed{181.7 \text{ g}} \)

(c)
Premium apples are in the range \( 170 < W < 185 \).
\( P(170 < W < 185) \approx 0.628364 \).
Answer: \( \boxed{62.8\%} \)

(d)
Let \( X \) be the number of premium apples in a box of 40.
\( X \sim B(40, p) \) where \( p \approx 0.628364 \).
We need \( P(X \geq 30) \). Using binomial CDF:
\( P(X \geq 30) = 1 – P(X \leq 29) \approx 0.073861 \).
Answer: \( \boxed{0.0739} \)

(e)
Let \( Y \) be the number of boxes satisfying the condition out of 10.
\( Y \sim B(10, 0.073861) \).
We need \( P(Y = 4) \).
\( P(Y = 4) \approx 0.003944 \).
Answer: \( \boxed{0.00394} \)

(f)
Let \( P(M < 170) = x \). Since \( P(M < 170) = 2 P(M > 185) \), then \( P(M > 185) = 0.5x \).
The total probability sums to 1:
\( x + 0.82 + 0.5x = 1 \implies 1.5x = 0.18 \implies x = 0.12 \).
So, \( P(M < 170) = 0.12 \) and \( P(M < 185) = 0.12 + 0.82 = 0.94 \).

Find the corresponding z-scores:
\( z_1 = \Phi^{-1}(0.12) \approx -1.17499 \)
\( z_2 = \Phi^{-1}(0.94) \approx 1.55477 \)
Set up the standardization equations:
1) \( \frac{170 – \mu}{\sigma} = -1.17499 \implies 170 – \mu = -1.17499\sigma \)
2) \( \frac{185 – \mu}{\sigma} = 1.55477 \implies 185 – \mu = 1.55477\sigma \)
Subtract equation (1) from (2):
\( 15 = (1.55477 – (-1.17499))\sigma = 2.72976\sigma \)
\( \sigma = \frac{15}{2.72976} \approx 5.49499 \).
Substitute \( \sigma \) back to find \( \mu \):
\( \mu = 170 + 1.17499(5.49499) \approx 176.456 \).
Answer: \( \boxed{176 \text{ g}} \)