A manufacturer uses a machine to make metal rods - Edexcel - A-Level Maths Statistics - Question 2 - 2022 - Paper 1

To find the proportion of metal rods between 7.94 cm and 8.09 cm, we need to calculate their corresponding z-scores.

For L = 7.94: $z_1 = \frac{7.94 - 8}{0.05} = -1.2$

For L = 8.09: $z_2 = \frac{8.09 - 8}{0.05} = 1.8$

Using the standard normal distribution table:

• For $z_1$ = -1.2, we find a cumulative probability of approximately 0.1151.
• For $z_2$ = 1.8, the cumulative probability is approximately 0.9641.

Thus, the proportion of metal rods between these lengths:

$P(7.94 < L < 8.09) = P(z_2) - P(z_1) = 0.9641 - 0.1151 = 0.849$

Therefore, the proportion is approximately 0.849.

To calculate the expected profit per 500 metal rods, we first consider the profit associated with each range of lengths:

• For L < 7.94, the profit is -15p (cost minus scrap value).
• For 7.94 ≤ L ≤ 8.09, the profit is 30p (selling price minus production cost).
• For L > 8.09, the profit is 30p (selling price minus production cost minus shortening cost).

Using previously calculated probabilities:

• The probability for L < 7.94 is 0.1151.
• The probability for 7.94 ≤ L ≤ 8.09 is 0.849.
• The probability for L > 8.09 is 0.035.

Calculating the expected profit: $ext{Expected Profit} = (0.1151 \times -15 + 0.849 \times 30 + 0.035 \times 30) \times 500$

This leads to: $= (0.1151 \times -15 + 0.849 \times 30 + 0.035 \times 30) \times 500 = 122 \text{pounds}$

To determine if the manufacturer is likely to achieve its aim for 95% of batches being accepted, we first calculate the probability of having fewer than 6 faulty hinges.

Let X represent the number of faulty hinges in a sample of 200. X follows a binomial distribution with parameters n = 200 and p = 0.015.

We need: $P(X < 6) = P(X \leq 5)$

Using normal approximation to the binomial: Mean is $np = 200 \times 0.015 = 3$ and variance is $np(1-p) = 200 \times 0.015 \times 0.985 = 2.955 $$

Using the continuity correction: $P(X \leq 5) \approx P(X < 5.5)$ Convert to z: $z = \frac{5.5 - 3}{\sqrt{2.955}} \approx 1.96$

Checking the z-table, we find: $P(Z < 1.96) \approx 0.975$

This means there is a 97.5% chance that fewer than 6 hinges are faulty, thus indicating the manufacturer is likely to achieve its aim.