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

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A manufacturer uses a machine to make metal rods. The length of a metal rod, L cm, is normally distributed with - a mean of 8 cm - a standard deviation of x cm Give... show full transcript

Worked Solution & Example Answer:A manufacturer uses a machine to make metal rods - Edexcel - A-Level Maths Statistics - Question 2 - 2022 - Paper 1

Step 1

show that x = 0.05 to 2 decimal places

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Answer

To find the value of x, we start with the concept of the z-score in a normal distribution. We know that:

P(L<7.902)=0.025P(L < 7.902) = 0.025

Using the z-table, the z-score corresponding to 2.5% is approximately -1.96. Therefore:

z=7.9028xz = \frac{7.902 - 8}{x}

Substituting the known values gives us:

1.96=7.9028x-1.96 = \frac{7.902 - 8}{x}

Solving for x, we have:

x=7.90281.96=0.05x = \frac{7.902 - 8}{-1.96} = 0.05

This confirms that x = 0.05 to 2 decimal places.

Step 2

Calculate the proportion of metal rods that are between 7.94 cm and 8.09 cm in length

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Answer

To calculate the proportion of metal rods between these lengths, we need to find:

P(7.94<L<8.09)P(7.94 < L < 8.09)

Calculating the z-scores for these lengths:

  1. For L = 7.94: z7.94=7.9480.05=1.2z_{7.94} = \frac{7.94 - 8}{0.05} = -1.2
  2. For L = 8.09: z8.09=8.0980.05=1.8z_{8.09} = \frac{8.09 - 8}{0.05} = 1.8

Using the z-table, we find the probabilities:

P(Z<1.2)0.1151P(Z < -1.2) \approx 0.1151 P(Z<1.8)0.9641P(Z < 1.8) \approx 0.9641

Thus, the proportion is:

P(7.94<L<8.09)=P(Z<1.8)P(Z<1.2)0.96410.1151=0.849P(7.94 < L < 8.09) = P(Z < 1.8) - P(Z < -1.2) \approx 0.9641 - 0.1151 = 0.849

Step 3

Calculate the expected profit per 500 of the metal rods

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Answer

First, we need to determine the potential profit from each range of metal rod lengths:

  1. For rods less than 7.94 cm:

    • Probability = 0.1151
    • Profit = 0.05 - 0.20 = -0.15
  2. For rods between 7.94 cm and 8.09 cm:

    • Probability = 0.849
    • Profit = 0.50 - 0.20 = 0.30
  3. For rods longer than 8.09 cm:

    • Probability = 0.0359
    • Profit after shortening = 0.50 - 0.20 - 0.10 = 0.20

Calculating profit per 500 rods:

  • Expected profit contribution from fewer than 7.94 cm: 500×(0.1151×0.15)=8.63500 \times (0.1151 \times -0.15) = -8.63
  • Expected profit contribution from between 7.94 cm and 8.09 cm: 500×(0.849×0.30)=127.35500 \times (0.849 \times 0.30) = 127.35
  • Expected profit contribution from more than 8.09 cm: 500×(0.0359×0.20)=3.59500 \times (0.0359 \times 0.20) = 3.59

Adding these contributions gives:

ExpectedProfit=8.63+127.35+3.59=122.31.Expected Profit = -8.63 + 127.35 + 3.59 = 122.31.

Thus, the expected profit for 500 rods is approximately £122.

Step 4

Explain whether the manufacturer is likely to achieve its aim

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Answer

The manufacturer's aim is for 95% of batches to be acceptable, which translates to a maximum of 6 faulty hinges in a sample of 200.

The probability of a hinge being faulty is 0.015. The expected number of faults in the sample can be calculated using:

E(X)=n×p=200×0.015=3E(X) = n \times p = 200 \times 0.015 = 3

Using the binomial probability formula, we can assess the likelihood of having fewer than 6 faulty hinges (i.e., P(X<6)P(X < 6)). Calculating P(X5)P(X \leq 5), we find:

P(X5)0.9176P(X \leq 5) \approx 0.9176

This suggests that there is a 91.76% chance of a batch being acceptable. Since this is less than the manufacturer's aim of 95%, it is unlikely that they will consistently achieve their goal.

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