A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm. Strips with length either less than 49 cm or greater than 50.75 cm cannot be used. Given that 2.5% of the cut lengths exceed 50.98 cm. (a) find the probability that a randomly chosen strip of metal can be used. Ten strips of metal are selected at random. (b) Find the probability fewer than 4 of these strips cannot be used. A second machine cuts strips of metal of length X cm, where X is normally distributed with standard deviation 0.6 cm. A random sample of 15 strips cut by this second machine was found to have a mean length of 50.4 cm. (c) Stating your hypotheses clearly and using a 1% level of significance, test whether or not the mean length of all the strips, cut by the second machine, is greater than 50.1 cm.

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A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm - Edexcel - A-Level Maths Statistics - Question 3 - 2017 - Paper 2

Question 3

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A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm. Strips with length either less than 49 cm or greater ... show full transcript

Worked Solution & Example Answer:A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm - Edexcel - A-Level Maths Statistics - Question 3 - 2017 - Paper 2

Step 1

(a) find the probability that a randomly chosen strip of metal can be used.

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Answer

To solve this part, we first need to determine the acceptable range of strip lengths. The lengths that cannot be used are those less than 49 cm and greater than 50.75 cm.

Using the standard normal distribution, we find the probability that a randomly chosen strip can be used:

Calculate the Z-score for 50.98 cm: $Z = \frac{50.98 - 50}{0.5} = 1.96$
The probability that L exceeds 50.98 has been given as 2.5%. Therefore, the probability that it can be used is: $P(L < 50.98) = 1 - P(L > 50.98) = 1 - 0.025 = 0.975$
Now, we calculate the total probability of a strip having a length within the acceptable range: $P(49 < L < 50.75) = P(L < 50.75) - P(L < 49)$ We can find:
- Calculate Z-score for 50.75: $Z_{50.75} = \frac{50.75 - 50}{0.5} = 1.5 \Rightarrow P(L < 50.75) \approx 0.9332$
- For 49: $Z_{49} = \frac{49 - 50}{0.5} = -2 \Rightarrow P(L < 49) \approx 0.0228$
Thus, placing these values in the formula: $P(49 < L < 50.75) = 0.9332 - 0.0228 = 0.9104$

Therefore, the probability that a randomly chosen strip can be used is approximately 0.910.

Step 2

(b) Find the probability fewer than 4 of these strips cannot be used.

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Answer

This part deals with binomial probability.

Let S be the number of strips that cannot be used. With n = 10 and p being the probability of choosing a strip that cannot be used. Since the acceptable length is from 49 to 50.75, we find:

Calculate p (probability of a strip not being useful): $p = 1 - P(49 < L < 50.75) = 1 - 0.910 = 0.090$
Therefore, S follows a Binomial distribution: S ~ B(10, 0.09).
We need to determine the probability that fewer than 4 strips cannot be used: $P(S < 4) = P(S = 0) + P(S = 1) + P(S = 2) + P(S = 3)$

Using the binomial probability formula: $P(S = k) = \binom{n}{k} p^k (1-p)^{n-k}$
- For k=0: $P(S=0) = \binom{10}{0} (0.09)^0 (0.91)^{10} \approx 0.409$
- For k=1: $P(S=1) = \binom{10}{1} (0.09)^1 (0.91)^9 \approx 0.360$
- For k=2: $P(S=2) = \binom{10}{2} (0.09)^2 (0.91)^8 \approx 0.141$
- For k=3: $P(S=3) = \binom{10}{3} (0.09)^3 (0.91)^7 \approx 0.032$
Combine these probabilities: $P(S < 4) \approx 0.409 + 0.360 + 0.141 + 0.032 = 0.942$

Thus, the probability that fewer than 4 strips cannot be used is approximately 0.942.

Step 3

(c) Stating your hypotheses clearly and using a 1% level of significance, test whether or not the mean length of all the strips, cut by the second machine, is greater than 50.1 cm.

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Answer

In this part, we need to perform a hypothesis test for the mean.

State the hypotheses:
- Null hypothesis (H0): ( ar{X} = 50.1 ) cm
- Alternative hypothesis (H1): ( ar{X} > 50.1 ) cm
Identify the sample data:
- Mean of the sample (( ar{X} )): 50.4 cm
- Standard deviation (( rac{0.6}{\sqrt{15}} )): 0.1559 cm
- Sample size (n): 15
Calculate the Z-score for the sample mean: $Z = \frac{\bar{X} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{50.4 - 50.1}{0.1559} \approx 1.93$
Find the p-value:
- Using Z-tables or standard normal distribution, we find the p-value for ( Z > 1.93 ): $p \approx 0.0264$
Decision and conclusion:
- Compare p-value with significance level (( \alpha = 0.01 )).
- Since 0.0264 > 0.01, we cannot reject the null hypothesis.

Therefore, there is insufficient evidence to conclude that the mean length of strips cut by the second machine is greater than 50.1 cm.

A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm - Edexcel - A-Level Maths Statistics - Question 3 - 2017 - Paper 2

Worked Solution & Example Answer:A machine cuts strips of metal to length L cm, where L is normally distributed with standard deviation 0.5 cm - Edexcel - A-Level Maths Statistics - Question 3 - 2017 - Paper 2

(a) find the probability that a randomly chosen strip of metal can be used.

(b) Find the probability fewer than 4 of these strips cannot be used.

(c) Stating your hypotheses clearly and using a 1% level of significance, test whether or not the mean length of all the strips, cut by the second machine, is greater than 50.1 cm.

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