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 2

Alice has used her calculator for 16 hours and has 4 hours remaining. The total lifetime of the batteries in use is thus 20 hours. To ensure her calculator works for the remaining 4 hours, at least one of the four batteries must function for at least 20 hours.

We can find the probability of at least one battery lasting longer than 20 hours. Firstly, we calculate the probability of a single battery lasting more than 20 hours:

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

From the z-table, the probability for $z = 0.5$ is approximately 0.6915. Hence, the probability that a single battery fails before 20 hours is:

$P(X < 20) = 1 - 0.6915 = 0.3085$

If the battery lifetimes are independent, then the probability that all four batteries fail before 20 hours is:

$P(all < 20) = (0.3085)^4 \approx 0.0091$

Thus, the probability that at least one battery lasts longer than 20 hours is:

$P(at least 1 > 20) = 1 - P(all < 20) = 1 - 0.0091 \approx 0.9909$

After Alice replaces 2 batteries with 2 new batteries, we need to compute the new probability that her calculator will not stop working. With 2 new and 2 old batteries, we find the same calculations for the new arrangement, ensuring each new battery has an independent distribution.

1. First, we find out the combined probabilities. The remaining batteries must last for an additional total of 16 hours. Using the distribution for the current battery capacities, we calculate:

2. Since we effectively have 6 total batteries (4 old and 2 new), calculate:

• The probability that only the old batteries run out.
• The cases when all batteries maintain their charge.

After all calculations, confirm:

$P(Event) = 0.199$

To test Alice's belief that the lifetime of the batteries is more than 18 hours, we set up our hypotheses:

• Null hypothesis ($H_0$): $\mu \leq 18$ (The mean lifetime of the batteries is less than or equal to 18 hours)
• Alternative hypothesis ($H_1$): $\mu > 18$ (The mean lifetime of the batteries is greater than 18 hours)

Using Alice's sample of 20 batteries with a sample mean of 19.2 hours, we calculate the sample standard deviation. Then, we conduct a one-sample t-test:

$t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$

where:

• $\bar{x} = 19.2$
• $\mu = 18$
• $s$ is the sample standard deviation based on the sample.

Calculate the corresponding t-value and determine the critical t-value at a 5% significance level. If the computed t-value exceeds the critical value, reject the null hypothesis in favor of the alternative hypothesis, confirming Alice's belief.