Question

x is a normally distributed random variable with mean 29 and standard deviation 8. what is the probability that x is between 12 and 46? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate z - scores for 12 and 46

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 29$ (mean) and $\sigma = 8$ (standard deviation).

For $x = 12$:
$z_1=\frac{12 - 29}{8}=\frac{- 17}{8}=-2.125$

For $x = 46$:
$z_2=\frac{46 - 29}{8}=\frac{17}{8}=2.125$

Step2: Find the probabilities corresponding to the z - scores

We need to find $P(-2.125

Using the standard normal distribution table (or a calculator with normal distribution functions), we know that:

$P(Z < 2.125)\approx0.9832$

$P(Z < - 2.125)=1 - P(Z < 2.125)\approx1 - 0.9832 = 0.0168$

Step3: Calculate the probability between the two z - scores

$P(-2.125

Answer:

0.966