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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 Giv... 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 use the standard normal distribution. The z-score corresponding to the lower 2.5% is -1.96 (from z-tables). Thus,

P(L<7.902)=0.025=P(L8x<1.96)P(L < 7.902) = 0.025 = P\left( \frac{L - 8}{x} < -1.96 \right)

Rearranging gives:

L8x=1.96    L=1.96x+8\frac{L - 8}{x} = -1.96 \implies L = -1.96x + 8

Substituting L = 7.902 into the equation, we have:

7.902=1.96x+87.902 = -1.96x + 8

Solving this gives:

1.96x=87.902=0.098 x=0.0981.96=0.051.96x = 8 - 7.902 = 0.098 \ \Rightarrow x = \frac{0.098}{1.96} = 0.05

Thus, it is shown that x = 0.05.

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 7.94 cm and 8.09 cm, we need the z-scores:

For L = 7.94:

Z=7.9480.05=1.2Z = \frac{7.94 - 8}{0.05} = -1.2

For L = 8.09:

Z=8.0980.05=1.8Z = \frac{8.09 - 8}{0.05} = 1.8

Using the z-table or calculator, we find:

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

Thus, the proportion:

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

To calculate the expected profit for 500 rods, we need to find the income from each category:

  1. For rods less than 7.94 cm:
    • Probability = 0.30
    • Selling price = 0.05
    • Profit = 0.05 - 0.20 = -0.15
  2. For rods between 7.94 cm and 8.09 cm:
    • Probability = 0.20
    • Selling price = 0.50
    • Profit = 0.50 - 0.20 = 0.30
  3. For rods longer than 8.09 cm:
    • Probability = 0.50
    • Shortening cost = 0.10
    • Selling price = 0.50
    • Profit = 0.50 - (0.20 + 0.10) = 0.20

Calculating expected profit:

E(Profit)=(0.30×0.15)+(0.20×0.30)+(0.50×0.20)E(Profit) = (0.30 \times -0.15) + (0.20 \times 0.30) + (0.50 \times 0.20)

= -0.045 + 0.06 + 0.10 = 0.115

For 500 rods, the expected profit:

ExpectedProfit=500×0.115=57.5pext(tothenearestpound:£1ext)Expected Profit = 500 \times 0.115 = 57.5 p \\ ext{(to the nearest pound: } £1 ext{)}

Step 4

Explain whether the manufacturer is likely to achieve its aim.

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Answer

Let X be the number of faulty hinges in a random sample of 200. The probability of a hinge being faulty, p = 0.015. Thus, np = 200 x 0.015 = 3 faulty hinges.

Using the binomial distribution, we can calculate:

P(X<6)=1P(X6)P(X < 6) = 1 - P(X \geq 6)

Using a normal approximation: Mean: μ = np = 3; Variance: σ² = np(1 - p) = 2.955. Standard deviation: σ = 1.721.

To find the probability: Convert to a z-score:

P(X \geq 6) = P\left(Z \geq \frac{6 - 3}{1.721} \right)$$ Calculating this results in a z-score of approximately 1.742, which corresponds to a value of about 0.0842. So, $$P(X < 6) = 1 - 0.0842 = 0.9158.

Since 0.9158 < 0.95, the manufacturer is unlikely to achieve its aim.

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