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

The total number of strips selected is 10, and we previously calculated that the probability of a strip not being usable is $1 - P(49 < L < 50.75) = 0.090$.

To find the probability that fewer than 4 strips cannot be used, use the binomial distribution: $P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$ where $X$ follows the distribution $X \sim B(n=10, p=0.090)$.

Calculate each probability: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$. Thus,

• For $k=0$: $P(X=0) = \binom{10}{0}(0.090)^0(0.910)^{10} \\ = 0.910^{10} \\approx 0.348$
• For $k=1$: $P(X=1) = \binom{10}{1}(0.090)(0.910)^{9} \\approx 0.383$
• For $k=2$: $P(X=2) = \binom{10}{2}(0.090)^2(0.910)^{8} \\approx 0.193$
• For $k=3$: $P(X=3) = \binom{10}{3}(0.090)^3(0.910)^{7} \\approx 0.056$

Therefore, $P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \\approx 0.348 + 0.383 + 0.193 + 0.056 = 0.980$

Hypothesis Statement:

• Null Hypothesis: $H_0: \mu = 50.1$ cm
• Alternative Hypothesis: $H_1: \mu > 50.1$ cm

Sample Mean and Standard Deviation:

• Sample mean: $\bar{X} = 50.4$ cm
• Standard deviation: $\sigma = 0.6$ cm, sample size $n = 15$.

Calculate the z-score: $z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} = \frac{50.4 - 50.1}{0.6 / \sqrt{15}} \\approx 1.936$

Determine the p-value and compare with significance level:

• For $z = 1.936$, the p-value is approximately $0.0264$.

Conclusion:

• Since $p = 0.0264 > 0.01$, we fail to reject the null hypothesis. Hence, there is insufficient evidence to conclude that the mean length of strips is greater than 50.1 cm.