The lifetime of Zaple smartphone batteries, $X$ hours, is normally distributed with mean 8 hours and standard deviation 1.5 hours - AQA - A-Level Maths Pure - Question 17 - 2020 - Paper 3

To find $P(6 < X < 10)$, we standardize the variable:

$Z = \frac{X - \mu}{\sigma}$

Where $\mu = 8$ and $\sigma = 1.5$. We calculate:

1. For $X = 6$: $Z_1 = \frac{6 - 8}{1.5} = -1.33$

2. For $X = 10$: $Z_2 = \frac{10 - 8}{1.5} = 1.33$

Using a standard normal distribution table, we find:

P(Z < -1.33) \approx 0.0918$$ Thus, $$P(6 < X < 10) = P(Z < 1.33) - P(Z < -1.33) = 0.9082 - 0.0918 = 0.8164$$ Therefore, $P(6 < X < 10) \approx 0.818$.

To find the 90th percentile, we need to find the value of $X$ such that:

$P(X < x) = 0.90$

Converting this value to the Z-score:

Using standard normal distribution tables, we find:

$Z \approx 1.28$

Now we convert this back to $X$ using the Z-formula:

$x = \mu + Z \sigma = 8 + 1.28(1.5) = 8 + 1.92 = 9.92$

Therefore, the lifetime exceeded by 90% of Zaple batteries is approximately 9.92 hours.

Given that 25% of Kaphone batteries last less than 5 hours, we set:

$P(Y < 5) = 0.25$

Standardizing this value:

Using the Z-score: $Z = \frac{5 - 7}{\tau}$ Using the standard normal distribution table, we find:

$Z \approx -0.6745$

Setting the equation:

$-0.6745 = \frac{5 - 7}{\tau} \Rightarrow -0.6745 \tau = -2 \Rightarrow \tau = \frac{2}{0.6745} \approx 2.97$

Thus, $\tau \approx 2.97$, correct to three significant figures.