The lifetime of Zaple smartphone batteries, $X$ hours, is normally distributed with mean 8 hours and standard deviation 1.5 hours. 17 (a) (i) Find $P(X eq 8)$ 17 (a) (ii) Find $P(6 < X < 10)$ 17 (b) Determine the lifetime exceeded by 90% of Zaple smartphone batteries. 17 (c) A different smartphone, Kaphone, has its battery's lifetime, $Y$ hours, modelled by a normal distribution with mean 7 hours and standard deviation $ au$. 25% of randomly selected Kaphone batteries last less than 5 hours. Find the value of $ au$, correct to three significant figures.

A-Level AQA Maths: Pure 2.12 Modelling with Functions: The lifetime of Zaple smartphone

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

Question 17

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The lifetime of Zaple smartphone batteries, $X$ hours, is normally distributed with mean 8 hours and standard deviation 1.5 hours. 17 (a) (i) Find $P(X eq 8)$ 17 ... show full transcript

Worked Solution & Example Answer: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

Step 1

Find $P(X eq 8)$

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Answer

To find the probability that a Zaple smartphone battery does not last exactly 8 hours, we recognize that in a continuous distribution, the probability of any single point is zero. Thus, we calculate:

eq 8) = 1 - P(X = 8) = 1$$ Therefore, $P(X eq 8) = 1$.

Step 2

Find $P(6 < X < 10)$

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Answer

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

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

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

For $X = 6$ : $Z_1 = \frac{6 - 8}{1.5} = -1.33$
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, $P(6 < X < 10) \approx 0.818$.

Step 3

Determine the lifetime exceeded by 90% of Zaple smartphone batteries.

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Answer

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.

Step 4

Find the value of $\tau$, correct to three significant figures.

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Answer

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.

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

Worked Solution & Example Answer: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

Find $P(X eq 8)$

Find $P(6 < X < 10)$

Determine the lifetime exceeded by 90% of Zaple smartphone batteries.

Find the value of $\tau$, correct to three significant figures.

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