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The random variable $X \sim N(\mu, 5^2)$ and $P(X < 23) = 0.9192$ - Edexcel - A-Level Maths Statistics - Question 2 - 2011 - Paper 2

Question 2

$The-random-variable-$X-\sim-N(\mu,-5^2)$-and-$P(X-<-23)-=-0.9192$-Edexcel-A-Level Maths Statistics-Question 2-2011-Paper 2.png$

The random variable $X \sim N(\mu, 5^2)$ and $P(X < 23) = 0.9192$. (a) Find the value of $\mu$. (b) Write down the value of $P(\mu < X < 23)$.

Worked Solution & Example Answer:The random variable $X \sim N(\mu, 5^2)$ and $P(X < 23) = 0.9192$ - Edexcel - A-Level Maths Statistics - Question 2 - 2011 - Paper 2

Step 1

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Answer

To find the value of $\mu$ , we start by standardizing the variable:

Using the z-score formula:
$z = \frac{x - \mu}{\sigma}$
where $x = 23$ and $\sigma = 5$ . Thus, we have: $\frac{23 - \mu}{5} = 1.40$ .
Rearranging gives:
$23 - \mu = 5 * 1.40$
$\Rightarrow 23 - \mu = 7$ .
Solving for $\mu$ :
$\mu = 23 - 7 = 16$ .

Thus, the value of $\mu$ is $16$ .

Step 2

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Answer

Given that $\mu$ is $16$ , we need to find:

$P(16 < X < 23)$ .

This can be rewritten using the standard normal distribution:

For $16$ : $z = \frac{16 - 16}{5} = 0$ .

For $23$ : $z = \frac{23 - 16}{5} = 1.4$ .

Thus, we can look up the values in the standard normal table or use a calculator to find:

$P(0 < Z < 1.4) = 0.4192$ .

Therefore, the value of $P(\mu < X < 23)$ is $0.4192$ .

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