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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

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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.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

Find the value of $\mu$.

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Answer

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

  1. Using the z-score formula:
    z=x−μσz = \frac{x - \mu}{\sigma}
    where x=23x = 23 and σ=5\sigma = 5. Thus, we have: 23−μ5=1.40\frac{23 - \mu}{5} = 1.40.

  2. Rearranging gives:
    23−μ=5∗1.4023 - \mu = 5 * 1.40
    ⇒23−μ=7\Rightarrow 23 - \mu = 7.

  3. Solving for μ\mu:
    μ=23−7=16\mu = 23 - 7 = 16.

Thus, the value of μ\mu is 1616.

Step 2

Write down the value of $P(\mu < X < 23)$.

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Answer

Given that μ\mu is 1616, we need to find:

P(16<X<23)P(16 < X < 23).

This can be rewritten using the standard normal distribution:

  • First, find the z-scores for μ\mu and 2323:

For 1616: z=16−165=0z = \frac{16 - 16}{5} = 0.

For 2323: z=23−165=1.4z = \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)=0.4192P(0 < Z < 1.4) = 0.4192.

Therefore, the value of P(μ<X<23)P(\mu < X < 23) is 0.41920.4192.

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