Question

question 10
given a normal distribution with a mean of 100 and a standard deviation of 10, what is the probability that
x > 79?
0.0179
0.9821
0.4821

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{X-\mu}{\sigma}\), where \(\mu = 100\), \(\sigma=10\), and \(X = 79\).

$$z=\frac{79 - 100}{10}=\frac{-21}{10}=- 2.1$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X>79)\), which is equivalent to \(P(Z>-2.1)\) in the standard normal distribution.
Since \(P(Z > z)=1 - P(Z\leq z)\) for a standard normal random variable \(Z\).
From the standard normal table, \(P(Z\leq - 2.1)=0.0179\)

$$P(Z>-2.1)=1 - 0.0179 = 0.9821$$

Answer:

0.9821