Question

for the standard normal distribution shown on the right, find the probability of z occurring in the indicated region. (round to four decimal places as needed.)
draw a normal curve with a mean of 58 and a standard deviation of 11. choose the correct graph of the normal curve below.
assume the random variable x is normally distributed with mean \\( \mu = 81 \\) and standard deviation \\( \sigma = 5 \\). find the indicated probability.
\\( p ( x < 71 ) = \\) (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Given \(\mu = 81\), \(\sigma=5\), and \(x = 71\).

$$z=\frac{71 - 81}{5}=\frac{-10}{5}=- 2$$

Step2: Find the probability using the standard normal table

We want to find \(P(X\lt71)\), which is equivalent to \(P(Z\lt - 2)\) (since \(Z=\frac{X-\mu}{\sigma}\)).
Looking up the value of \(z=-2\) in the standard normal table (the cumulative - distribution function of the standard normal distribution \(\varPhi(z)\)), we know that \(P(Z\lt - 2)=0.0228\)

Answer:

\(0.0228\)