a) Use complement rule:

\( P(82.4 < H < 87.3) = 1 – P(H < 82.4) – P(H > 87.3) \)

\( = 1 – 0.213 – 0.409 = 0.378 \) [2]

b) Find \( z \)-scores:

\( P(Z < -0.80) \approx 0.2119 \approx 0.213 \implies z_1 = -0.80 \)

\( P(Z > 0.23) \approx 0.4091 \approx 0.409 \implies z_2 = 0.23 \)

\( z = \frac{H – \mu}{\sigma} \)

For \( H = 82.4 \):

\( -0.80 = \frac{82.4 – \mu}{\sigma} \) …(1)

For \( H = 87.3 \):

\( 0.23 = \frac{87.3 – \mu}{\sigma} \) …(2)

From (1):

\( \mu = 82.4 + 0.80 \times \sigma \) …(3)

From (2):

\( 87.3 – \mu = 0.23 \times \sigma \)

Substitute (3):

\( 87.3 – (82.4 + 0.80 \times \sigma) = 0.23 \times \sigma \)

\( 4.9 = 1.03 \times \sigma \)

\( \sigma \approx \frac{4.9}{1.03} \approx 4.76 \)

Substitute \( \sigma \approx 4.76 \):

\( \mu \approx 82.4 + 0.80 \times 4.76 \approx 82.4 + 3.808 \approx 86.208 \approx 86.2 \) [5]

ci) Use binomial distribution:

\( n = 100 \), \( p = 0.409 \)

\( P(X = 32) = \binom{100}{32} \times 0.409^{32} \times 0.591^{68} \)

Using numerical tools:

\( P(X = 32) \approx 0.0158 \) [3]

cii) Use conditional probability:

\( P(X = 32 | X < 44) = \frac{P(X = 32)}{P(X < 44)} \)

\( P(X = 32) \approx 0.0158 \)

\( P(X < 44) \approx 0.703 \) (via binomial cumulative distribution)

\( P(X = 32 | X < 44) \approx \frac{0.0158}{0.703} \approx 0.0225 \) [3]

d) Use interquartile range for normal distribution:

\( \text{IQR} = Q_3 – Q_1 \approx 1.349 \times d \)

Given \( \text{IQR} = 4.52 \):

\( d \approx \frac{4.52}{1.349} \approx 3.35 \) [3]