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 Mechanics - Question 17 - 2020 - Paper 3

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

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

Where $\mu = 8$ and $\sigma = 1.5$.

1. For $X = 6$: $Z_6 = \frac{6 - 8}{1.5} = -1.33$

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

Now, we will consult the z-table: $P(Z < -1.33) \approx 0.0918$ $P(Z < 1.33) \approx 0.9082$

Thus,

$P(6 < X < 10) = P(Z < 1.33) - P(Z < -1.33) \approx 0.9082 - 0.0918 = 0.8164$

To find the lifetime that exceeds 90% of Zaple smartphone batteries, we need to determine the 90th percentile (Z value) for a standard normal distribution.

From the Z-table, the Z value for 0.90 is: $Z \approx 1.28$

Now we can find the corresponding lifetime: $X = \mu + Z \sigma = 8 + (1.28)(1.5) \approx 9.92 \text{ hours}$

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

Given that 25% of Kaphone batteries last less than 5 hours, we first need to find the corresponding Z value: $Z \approx -0.6745$

Now we can set the equation for the normal variable: $5 = 7 + Z \sigma$

Substituting the Z value: $5 = 7 - 0.6745 \sigma$

Rearranging gives: $\sigma = \frac{7 - 5}{0.6745} \approx 2.96$

Therefore, the value of $\sigma$, correct to three significant figures, is: $\sigma \approx 2.96$