Question

x is a normally distributed random variable with mean 50 and standard deviation 21.
what is the probability that x is between 4 and 96?
write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate z-scores for 4 and 96

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 50$ (mean) and $\sigma=21$ (standard deviation).

For $x = 4$:
$z_1=\frac{4 - 50}{21}=\frac{- 46}{21}\approx - 2.1905$

For $x = 96$:
$z_2=\frac{96 - 50}{21}=\frac{46}{21}\approx2.1905$

Step2: Find the probabilities corresponding to the z - scores

We need to find $P(-2.1905

Using the standard normal distribution table (or a calculator with normal distribution functions):

$P(Z < 2.1905)\approx0.9857$

$P(Z < - 2.1905)=1 - P(Z < 2.1905)\approx1 - 0.9857 = 0.0143$

Step3: Calculate the probability between the two z - scores

$P(-2.1905

Answer:

0.971