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

Let $p = 0.0668$. We can use the binomial distribution with $n = 5$ to calculate the probability of at least two rods.

First, we find the probabilities of 0 and 1 successful outcomes:

1. Probability of exactly 0 rods:
   $P(X = 0) = {5 \choose 0} p^0 (1-p)^5 = (0.9332)^5 \approx 0.7059$

2. Probability of exactly 1 rod:
   $P(X = 1) = {5 \choose 1} p^1 (1-p)^4 = 5(0.0668)(0.9332)^4 \approx 0.2587$

Therefore, the probability of at least 2 rods is:

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.7059 - 0.2587 = 0.0354$

To conduct the hypothesis test:

Null Hypothesis ($H_0$): $\mu = 40$ mm

Alternative Hypothesis ($H_1$): $\mu \neq 40$ mm

Using the sample data:

• Sample = [39.5, 40.0, 39.7, 40.2, 39.8, 40.1, 39.6]
• Mean of the sample ($\bar{x}$) = 39.87 mm
• Rough standard deviation ($\sigma_t$) = $\frac{0.2}{\sqrt{10}} \approx 0.0632455532$
• Calculate the Z-score:

$Z = \frac{\bar{x} - \mu}{\sigma_t} = \frac{39.87 - 40}{0.0632455532} \approx -2.05$

Critical value for the test at 5% significance level: $Z_{crit} = \pm 1.96$

Since $-2.05 < -1.96$, we reject the null hypothesis at the 5% significance level and conclude that the machine setting has become inaccurate.