a) From the cumulative frequency diagram, the median weight is \( 50 \) (g) [1]

bi) From the diagram, 65 rats weigh less than 70 grams.

Calculate percentage using the formula: \( \frac{65}{80} \times 100 = 81.25 \) (%) [3]

bii) Total frequency is 80, with known values 45 + \( q \) + 5, and from the diagram, \( p = 10 \) [2]

biii) Total frequency is 80, with \( p = 10 \), 45, and 5 given, so \( q = 80 – 10 – 45 – 5 = 20 \) [2]

c) Use mid-interval values: \( 15, 45, 75, 105 \) for intervals \( 0-30, 30-60, 60-90, 90-120 \).

Mean \( \overline{x} = 52.5 \) (exact), standard deviation \( \sigma = 22.5 \) (exact) [3]

d) Using normal distribution with \( \mu = 52.5 \), \( \sigma = 22.5 \), the percentage is \( 78.2 \) (%) [2]

e) Model as binomial \( X \sim \text{B}(5, 0.782) \), where 0.782 is derived from part d).

Calculate \( P(X \leqslant 3) \):

\( P(X = 0) = \binom{5}{0} \times (0.782)^0 \times (1 – 0.782)^5 = 1 \times 1 \times (0.218)^5 \)

\( (0.218)^5 \approx 0.000228 \)

\( P(X = 0) \approx 0.000228 \)

\( P(X = 1) = \binom{5}{1} \times (0.782)^1 \times (0.218)^4 \)

\( \binom{5}{1} = 5 \), \( (0.218)^4 \approx 0.00228 \), \( 5 \times 0.782 \times 0.00228 \approx 0.00892 \)

\( P(X = 1) \approx 0.00892 \)

\( P(X = 2) = \binom{5}{2} \times (0.782)^2 \times (0.218)^3 \)

\( \binom{5}{2} = 10 \), \( (0.782)^2 \approx 0.6115 \), \( (0.218)^3 \approx 0.01036 \), \( 10 \times 0.6115 \times 0.01036 \approx 0.0634 \)

\( P(X = 2) \approx 0.0634 \)

\( P(X = 3) = \binom{5}{3} \times (0.782)^3 \times (0.218)^2 \)

\( \binom{5}{3} = 10 \), \( (0.782)^3 \approx 0.478 \), \( (0.218)^2 \approx 0.0475 \), \( 10 \times 0.478 \times 0.0475 \approx 0.2272 \)

\( P(X = 3) \approx 0.2272 \)

\( P(X \leqslant 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \)

\( \approx 0.000228 + 0.00892 + 0.0634 + 0.2272 \approx 0.2997 \)

Rounding to three decimal places, \( P(X \leqslant 3) = 0.301 \) [3]