Question

x is a normally distributed random variable with mean 24 and standard deviation 25. what is the probability that x is between 7 and 32? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate z - scores for 7 and 32

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 24$ (mean) and $\sigma=25$ (standard deviation).

For $x = 7$:
$z_1=\frac{7 - 24}{25}=\frac{- 17}{25}=- 0.68$

For $x = 32$:
$z_2=\frac{32 - 24}{25}=\frac{8}{25}=0.32$

Step2: Find the probabilities corresponding to the z - scores

We use the standard normal distribution table (or z - table) to find $P(Z < z_1)$ and $P(Z < z_2)$.

From the z - table, $P(Z < - 0.68)=0.2483$ and $P(Z < 0.32) = 0.6255$.

Step3: Calculate the probability that X is between 7 and 32

The probability $P(7

Substitute the values: $0.6255 - 0.2483 = 0.3772$

Answer:

0.377