Question

x is a normally distributed random variable with mean 87 and standard deviation 19.
what is the probability that x is between 81 and 93?
write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate z-scores for 81 and 93

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 87$ (mean) and $\sigma=19$ (standard deviation).

For $x = 81$:
$z_1=\frac{81 - 87}{19}=\frac{- 6}{19}\approx - 0.316$

For $x = 93$:
$z_2=\frac{93 - 87}{19}=\frac{6}{19}\approx0.316$

Step2: Find the probabilities corresponding to the z - scores

We use the standard normal distribution table (or a calculator with normal distribution functions) to find $P(Z < z_1)$ and $P(Z < z_2)$.

From the standard normal table, $P(Z < - 0.316)\approx0.376$ and $P(Z < 0.316)\approx0.624$

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(- 0.316

Answer:

0.248