A factory manufactures aluminium rods - Leaving Cert Mathematics - Question 9A - 2010

Using the result from part (a), we know that:

• $p = 0.0668$ (the probability of selecting a rod less than 39.7 mm)
• $q = 1 - p = 0.9332$ (the probability of selecting a rod 39.7 mm or longer)

For a binomial distribution with $n = 5$, we want to find:

$P(X \geq 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)]$

Using the binomial formula:

$P(X = k) = \binom{n}{k} p^k q^{n-k}$

We compute the individual probabilities:

• For $k = 0$: $P(X = 0) = \binom{5}{0} (0.0668)^0 (0.9332)^5 \approx 0.6634$

• For $k = 1$: $P(X = 1) = \binom{5}{1} (0.0668)^1 (0.9332)^4 \approx 0.2865$

Now summing them up:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.6634 + 0.2865 \approx 0.9499$

Finally, we find:

$P(X \geq 2) = 1 - 0.9499 \approx 0.0501$

Therefore, the probability that at least two of them are less than 39.7 mm in length is approximately 0.0501.

To conduct the hypothesis test, we begin by establishing our null and alternative hypotheses:

• Null Hypothesis (H0): $\mu = 40$ mm (the machine's setting is accurate)
• Alternative Hypothesis (H1): $\mu \neq 40$ mm (the machine's setting is inaccurate)

Next, we gather the sample data:

Sample lengths: 39.5, 40.0, 39.7, 40.2, 39.8, 39.7, 40.2, 39.9, 40.1, 39.6

We calculate the sample mean ($\bar{x}$):

$\bar{x} = \frac{39.5 + 40.0 + 39.7 + 40.2 + 39.8 + 39.7 + 40.2 + 39.9 + 40.1 + 39.6}{10} = 39.87$

Next, we calculate the standard deviation of the sample mean:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.2}{\sqrt{10}} \approx 0.0632$

Now, we compute the Z-score for our sample mean:

$Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{39.87 - 40}{0.0632} \approx -2.07$

We compare this Z-score to the critical values at a 5% significance level (Z-critical values are approximately ±1.96 for a two-tailed test):

Since $-2.07 < -1.96$, we reject the null hypothesis.

Conclusion: At the 5% level of significance, we reject the null hypothesis, indicating that the machine's setting has become inaccurate.