A manufacturer uses a machine to make metal rods - Edexcel - A-Level Maths Statistics - Question 2 - 2022 - Paper 1
Question 2
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
Gi... 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 show that x=0.05, we need to use the normal distribution formula.
Given P(L<7.902)=0.025, we can find x using the z-score formula:
z = rac{L - ext{mean}}{ ext{standard deviation}}
Plugging in our values:
z = rac{7.902 - 8}{x} = -1.96
Solving this gives:
7.902−8=−1.96xx = rac{8 - 7.902}{1.96} = 0.05000
Thus, rounding to 2 decimal places gives 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 find the proportion of metal rods between 7.94 cm and 8.09 cm, we again use the standard normal distribution.
First, we find the z-scores for both lengths:
Now we consult the standard normal distribution table to find the probabilities:
= 0.9641 - 0.1151 = 0.8490$$
Thus, the proportion is approximately 0.849.
Step 3
Calculate the expected profit per 500 of the metal rods.
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Answer
To calculate the expected profit per 500 metal rods, we first determine the income from selling each category:
For L<7.94:
Proportion: 0.025, Profit: 0.05−0.20=−0.15 (Loss)
For 7.94extto8.09:
Proportion: 0.849, Profit: 0.50−0.20=0.30
For L>8.09:
Proportion: 0.126, Profit: (0.50−0.20−0.10)=0.20
The expected profit for 500 rods is thus:
extExpectedprofit=500imes(0.025imes−0.15+0.849imes0.30+0.126imes0.20)
Calculating this gives:
= 500 imes (0.0247) = 122.05\
The final expected profit is approximately £122.
Step 4
Explain whether the manufacturer is likely to achieve its aim.
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Answer
To determine if the manufacturer is likely to achieve the aim of 95% acceptance of hinges, we can use the binomial distribution as the probability of having a faulty hinge is given:
The probability of a hinge being faulty is p=0.015. We want to find the probability of getting more than 6 faulty hinges out of 200:
By using the binomial cumulative distribution:
P(Xextallowsacceptance)=P(X<6)
Using a binomial table or calculator:
P(X<6)=0.9176
This indicates that the likelihood of acceptance is about 91.76%, which is less than the desired 95%. Therefore, the manufacturer is unlikely to achieve their aim.