The lifetime, L, hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours - Edexcel - A-Level Maths Mechanics - Question 5 - 2018 - Paper 1

Given that Alice has used her calculator for 16 hours, she has 4 hours of exams left. To find the probability that her calculator will last the remaining 4 hours, we first need to calculate the probability of one battery lasting more than 20 hours:

$P(L > 20) = P(L > 16 | L > 16) = P(L > 20 | L > 16)$

We can use the Z-score for this:

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

Using the normal distribution table, we get:

$P(L < 20) \approx 0.6915$

Then,

$P(L > 20) = 1 - P(L < 20) \approx 1 - 0.6915 = 0.3085$

Since Alice's calculator runs on 4 batteries, we apply:

$P(L > 20)^4 \approx (0.3085)^4 \approx 0.0094$ (for her calculator not stopping for the remaining exams)

After the first 16 hours of exams, Alice has two new batteries. The probability of her calculator not stopping after replacing the chosen batteries:

We calculate:

$P = 1 - P(L < 16) \implies P(L > 16)$

We already calculated this as 0.6915. Thus:

$P(L > 16 | \text{2 new batteries}) = 0.6915 \cdot P(2 \text{ selected}) \approx 0.199$

This final result of 0.199 to three significant figures matches the expected answer.

The hypotheses can be formulated as follows:

• Null Hypothesis (H0): (\mu \leq 18) (the mean lifetime is 18 hours or less)
• Alternative Hypothesis (H1): (\mu > 18) (the mean lifetime is greater than 18 hours)

Using a sample of n = 20 batteries, with a mean (\bar{x} = 19.2) and a standard deviation estimated from the original normal distribution, we calculate the Z-score:

$Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \approx \frac{19.2 - 18}{4 / \sqrt{20}}$

This Z-score is compared against the critical value from Z-table for a significance level of 0.05:

Since the calculated Z-value exceeds the critical Z-value, we reject the null hypothesis, supporting Alice's belief.