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x is a normally distributed random variable with mean 24 and standard d…

Question

x is a normally distributed random variable with mean 24 and standard deviation 9. what is the probability that x is between 23 and 25? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Recall the Z - score formula

The Z - score formula is $Z=\frac{X - \mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $X$ is the value from the random variable. Given that $\mu = 24$, $\sigma=9$. We need to find the Z - scores for $X = 23$ and $X = 25$.

For $X = 23$:
$Z_1=\frac{23 - 24}{9}=\frac{- 1}{9}\approx - 0.111$

For $X = 25$:
$Z_2=\frac{25 - 24}{9}=\frac{1}{9}\approx0.111$

Step2: Use the standard normal distribution table (or calculator)

We want to find $P(23\lt X\lt25)=P(- 0.111\lt Z\lt0.111)$.

We know that $P(-a\lt Z\lt a)=2P(0\lt Z\lt a)$ (by the symmetry of the standard normal distribution).

First, find $P(Z\lt0.111)$ and $P(Z\lt - 0.111)$.

Using a standard normal calculator or table:
$P(Z\lt0.111)\approx0.544$
$P(Z\lt - 0.111)\approx0.456$

Then $P(-0.111\lt Z\lt0.111)=P(Z\lt0.111)-P(Z\lt - 0.111)=0.544 - 0.456 = 0.088$

Answer:

0.088