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, the lifetime of the batteries used in Alice’s calculator must be considered. Each battery has a remaining life:

$X' = L - 16$

We seek the probability that at least one battery lasts beyond the additional 4 hours needed:

Using previously calculated:

$P(L > 20) = 1 - P(L < 20) = 1 - 0.44621 \approx 0.55379$

Thus,

• Probability calculator works for remaining exams:

$P(L > 20 | L > 16) = \frac{P(L > 20)}{P(L > 16)} \approx \frac{0.55379}{0.6914} \approx 0.8012$

To find the probability that her calculator won’t stop working:

Let’s denote:

• $X_1, X_2$ as the lifetime of the selected new batteries

Then:

$P(L > 4 | L > 16) = P(L > 20 | L > 16)^2\times P(L > 20 | L > 16)$

From previous calculations,

$P(L > 4) \equiv 0.99776$ $\Rightarrow P(L > 20 | L > 16) \approx 0.44621$

Thus, the probability that the calculator will not stop working for the remaining exams:

$0.199 \text{ to 3 significant figures.}$

We test:

• Null Hypothesis: $H_0: \mu \leq 18$ (the mean lifetime is 18 hours or less)
• Alternative Hypothesis: $H_a: \mu > 18$ (the mean lifetime is more than 18 hours)

Using a one-sample Z-test:

$Z = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}}$

where:

• $\bar{X} = 19.2$ (sample mean)
• $\mu_0 = 18$ (hypothesized mean)
• $\sigma = ext{ Standard deviation} ext{ (assumed)}
• $n = 20$.

Calculating:

$Z = \frac{19.2 - 18}{4/\sqrt{20}} = \frac{1.2}{0.8944} \approx 1.34$

At a 5% significance level, the critical Z-value is approximately 1.645. Since $1.34 < 1.645$, we do not reject the null hypothesis. There is insufficient evidence to support Alice's belief that the mean lifetime is more than 18 hours.