Question

A random variable X is distributed normally with E(X) = 8 and σ(X) =3 . Find P(9≤X<11).

Answer 1

Answer on Question #43457 – Math - Statistics and Probability

A random variable $X$ is distributed normally with $E(X) = 8$ and $\sigma(X) = 3$. Find $P(9 \leq X < 11)$.

Solution

$X$ is normally distributed with parameters $\mu = E(X) = 8, \sigma = \sigma(X) = 3$.

If $X$ is distributed normally $N(\mu, \sigma^2)$, then $P(a \leq X < b) = P\left(\frac{a - \mu}{\sigma} \leq Z < \frac{b - \mu}{\sigma}\right)$, where $Z$ has the standard normal distribution.

The standardized variable is $Z = \frac{X - \mu}{\sigma} = \frac{X - 8}{3}$.

In particular,

$$\begin{array}{l} x = 9 \text{ gives } z = \frac{x - \mu}{\sigma} = \frac{9 - 8}{3} = \frac{1}{3} \approx 0.33, \\ x = 11 \text{ gives } z = \frac{x - \mu}{\sigma} = \frac{11 - 8}{3} = 1. \end{array}$$

To find $P(Z < 1) = 0.8413$ and $P(Z < 0.33) = 0.6293$, we use statistical tables or software.

Therefore, the required probability is

$$P(9 \leq X < 11) = P(0.33 \leq Z < 1) = P(Z < 1) - P(Z < 0.33) = 0.8413 - 0.6293 = 0.212.$$

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