(i) To find the probability that an apple can be used as a toffee apple, we calculate the probability that the diameter is between 4.1 cm and 5 cm. We standardize these values using the formula for the z-score:

\(z_1 = \frac{4.1 - 5.7}{0.8} = -2\)

\(z_2 = \frac{5 - 5.7}{0.8} = -0.875\)

The probability is given by:

\(P(4.1 < d < 5) = P(z < -0.875) - P(z < -2)\)

Using standard normal distribution tables:

\(\Phi(-0.875) = 0.1908\)

\(\Phi(-2) = 0.0228\)

Thus, \(P(4.1 < d < 5) = 0.1908 - 0.0228 = 0.168\)

(ii) For 250 apples, the number that can be used as toffee apples follows a binomial distribution with \(n = 250\) and \(p = 0.168\). We approximate this with a normal distribution:

\(np = 250 \times 0.168 = 42\)

\(npq = 34.944\)

Standard deviation \(\sigma = \sqrt{34.944} = 5.911\)

We find \(P(X < 50)\) using continuity correction:

\(P(X < 50) = P\left( z < \frac{49.5 - 42}{5.911} \right) = P(z < 1.2687)\)

Using standard normal distribution tables:

\(\Phi(1.2687) = 0.898\)