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

After using the calculator for 16 hours, Alice has 4 hours left for her exams. For her calculator not to stop working, at least 3 out of the 4 batteries must last longer than an additional 4 hours. We need to find:

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

Using the previously calculated value, we have:

$P(L > 16) = 0.6915$

Now for 4 batteries, the probability of at least 3 surviving is computed using the binomial distribution:

P(X ext{ successes}) = {n race k} p^k (1-p)^{n-k}

Thus, summing the probabilities, we arrive at:

$P(X ext{ at least } 3) = P(X = 3) + P(X = 4)$.

Calculating these, we determine the calculator's probability of continuing to function.

After replacing the used batteries, we calculate:

• The probability of drawing new batteries which ensures functionality is determined from prior calculations, ultimately leading us to:

$P(L > 20 | L > 16) = 0.4462$

We find that 2 batteries must operate beyond 20 hours:

$P( ext{2 new batteries functioning beyond remaining hours}) = 0.199.$

We set our hypotheses as follows:

• Null Hypothesis ($H_0$): The mean lifetime of the batteries is equal to 18 hours, H_0: eta = 18.
• Alternative Hypothesis ($H_a$): The mean lifetime is greater than 18 hours, H_a: eta > 18.

Using the sample mean of 19.2 hours from a sample size of 20, we compute the test statistic:

z = rac{ar{x} - eta}{s/ ext{sqrt}(n)} = rac{19.2 - 18}{4/ ext{sqrt}(20)}

After evaluating the z-value against critical z-values of alpha = 0.05, we can conclude that there is not enough evidence to reject $H_0$.