A car from the economy fleet is chosen at random - Leaving Cert Mathematics - Question 4 - 2015

This part requires using the binomial distribution, where the success probability $p = 0.734$ (probability from part (a)). We set up the following:

Using the binomial probability formula: $P(X = k) = {n \choose k} p^k (1 - p)^{n - k}$

Where:

• $n = 20$ (total cars)
• $k$ is the number of successes (cars lasting at least 40,000 km)

Calculating for:

• $P(20 \text{ successes}) = {20 \choose 20} (0.734)^{20} (0.266)^{0} = 0.0021$
• $P(19 \text{ successes}) = {20 \choose 19} (0.734)^{19} (0.266)^{1} = 0.0149$
• $P(18 \text{ successes}) = {20 \choose 18} (0.734)^{18} (0.266)^{2} = 0.0514$

Now, adding these probabilities together: $P(\text{at least } 18 \text{ successes}) = 0.0021 + 0.0149 + 0.0514 = 0.0684$(approximately).

Therefore, the probability that at least 18 out of 20 cars will last for 40,000 km is approximately $0.0684$.

For this hypothesis test, we set the null hypothesis as:
$H_0: \mu = 45000$
Against the alternative hypothesis:
$H_a: \mu \neq 45000$

We have a sample mean of $x = 43850$ km, sample size $n = 25$, and the population standard deviation $\sigma = 8000$ km, so we calculate the z-score as:

$z = \frac{x - \mu}{\sigma / \sqrt{n}} = \frac{43850 - 45000}{8000 / \sqrt{25}} = \frac{-1150}{1600} = -0.71875$

The critical z-value for a two-tailed test at the 5% level of significance is approximately $\pm 1.96$.

Since $-1.96 < -0.71875 < 1.96$, we fail to reject the null hypothesis.
Thus, the company cannot conclude that SafeRun tyres have a different lifespan than Everthread tyres.