The lifetime, L hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours - Edexcel - A-Level Maths Statistics - Question 5 - 2018 - Paper 2

After 16 hours, Alice has 4 batteries that have already been used. The probability that a battery lasts longer than 20 hours can be found using the Z-score again:

$Z = \frac{20 - 18}{4} = 0.5$

Thus,

$P(L > 20) = P(Z > 0.5) \approx 0.308537.$

Now, since she has 4 batteries, the probability that none of them fail is:

$P(L > 20)^4 = (0.308537)^4 \approx 0.036.$

Therefore, the probability that her calculator will not stop working for the remaining exams is approximately 0.036.

After the first 16 hours, the situation changes as she replaces 2 batteries. The calculation involves the probabilities of the new set of batteries:

From the 2 batteries, the probability they both last longer than 4 hours must be calculated:

$P(L > 4) = 1 - P(L \leq 4)$

Calculate this, and considering the 2 batteries:

$P(L > 4) * P(L > 20 | L > 16) = (0.99976) * (0.44621)^2 \approx 0.199.$

Let the null hypothesis be: $H_0: \mu \leq 18$ And the alternative hypothesis: $H_1: \mu > 18$

Using a sample size of 20 and a sample mean of 19.2 hours, we compute the test statistic: $Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} = \frac{19.2 - 18}{4 / \sqrt{20}} \approx 2.607.$

The critical value for a one-tailed test at the 5% significance level is approximately 1.645. Since $Z > 1.645$, we reject the null hypothesis, supporting Alice's belief that the mean lifetime is greater than 18 hours.