Question

x is a normally distributed random variable with mean 34 and standard deviation 9. what is the probability that x is between 31 and 37? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the dataset.

For $x = 31$, $\mu=34$, $\sigma = 9$:
$z_1=\frac{31 - 34}{9}=\frac{- 3}{9}\approx - 0.333$

For $x = 37$, $\mu = 34$, $\sigma=9$:
$z_2=\frac{37 - 34}{9}=\frac{3}{9}\approx0.333$

Step2: Find the probabilities corresponding to the z - scores

We use the standard normal distribution table (or a calculator with a normal - distribution function) to find $P(Z

From the standard normal table, $P(Z < - 0.333)\approx0.3694$ and $P(Z < 0.333)\approx0.6306$

Step3: Calculate the probability between the two z - scores

The probability that $Z$ is between $z_1$ and $z_2$ is $P(z_1

So, $P(-0.333

Answer:

0.261