Question

x is a normally distributed random variable with mean 3 and standard deviation 17. what is the probability that x is between 10 and 98? 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 original distribution. Given that $\mu = 3$, $\sigma=17$. We need to find the Z - scores for $X = 10$ and $X = 98$.

Step2: Calculate the Z - score for $X = 10$

Substitute $X = 10$, $\mu=3$, and $\sigma = 17$ into the Z - score formula:
$Z_1=\frac{10 - 3}{17}=\frac{7}{17}\approx0.4118$

Step3: Calculate the Z - score for $X = 98$

Substitute $X = 98$, $\mu = 3$, and $\sigma=17$ into the Z - score formula:
$Z_2=\frac{98 - 3}{17}=\frac{95}{17}\approx5.5882$

Step4: Use the standard normal distribution table (or calculator)

We want to find $P(10

For a standard normal distribution, as $z$ becomes very large (greater than 3 or 4), $\Phi(z)\approx1$. So $\Phi(5.5882)\approx1$.

We know that $\Phi(0.41)\approx0.6591$ and $\Phi(0.4118)$ can be approximated. Using a calculator or more precise table, $\Phi(0.4118)\approx0.6597$.

So $P(0.4118 < Z<5.5882)=1 - 0.6597 = 0.3403$ (approximate value, more precisely, using a calculator for the Z - scores:

Using a calculator, for $Z_1=\frac{10 - 3}{17}\approx0.4118$, $\Phi(0.4118)\approx0.6597$

For $Z_2=\frac{98 - 3}{17}\approx5.5882$, $\Phi(5.5882)\approx1$

So $P(10 < X<98)=P(0.4118 < Z<5.5882)=1 - 0.6597=0.3403$ (rounded to the nearest thousandth)

Answer:

0.340