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's calculator requires a total of 4 batteries. Since she has already used it for 16 hours and has 4 more hours of exams to sit, we need to calculate the probability that none of the remaining batteries fail during this time.

Assuming the remaining batteries have a mean life of 18 hours, we can denote the remaining time as:

• $T = 4$ hours.

We will find the probability that each of the 2 remaining batteries lasts more than 4 hours. Using the Z-score:

$Z = \frac{4 - 18}{4} = -3.5$

Then, checking the Z-table: $P(Z > -3.5) \approx 0.9998$

Since we must have both batteries last this duration: $P(X_1 > 4 \text{ and } X_2 > 4) = P(X > 4)^2 = (0.9998)^2 \approx 0.9996$

Thus, the probability that her calculator will not stop working during the remaining exams is approximately 0.9996.

In this part, after replacing the batteries, we need to repeat the probability calculation. Assuming that 2 batteries are replaced:

• After 16 hours with 2 batteries remaining:

Let the probability of one battery working during the next 4 hours be calculated similarly:

Assume they last longer than 4 hours. Since both new batteries have mean life over 18 hours: Using: $P(X > 4) = 0.9998$

We can represent the failure probabilities and apply: $P(X \leq 4) = 1 - 0.9998 = 0.0002$

We can show: $P(Failure) = 1 - P(Not stopping) = 0.199$

So, simplified, the probability that her calculator will not stop working for remainder of her exam is approximately 0.199.

Alice's belief can be tested using hypothesis testing. We set our hypotheses as follows:

• Null Hypothesis ($H_0$): The mean lifetime of the batteries is $\mu \leq 18$ hours.
• Alternative Hypothesis ($H_1$): The mean lifetime of the batteries is $\mu > 18$ hours.

Using a sample mean ($\bar{x}$) of 19.2 hours and a sample size (n) of 20:

• The sample standard deviation (assuming equal variance), $s$.
• We use the Z-test for means:

Calculate the test statistic: $Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$

After calculating the Z-value, we should compare it with the critical Z-value for a one-tailed test at $\\alpha = 0.05$. If the calculated Z-value exceeds the critical value, we reject $H_0$ in favor of $H_1$. If not, we fail to reject $H_0$. This allows us to validate or refute Alice's belief concerning battery life.