Use technology to solve the following problem: A normal population has a mean $\mu = 31$ and standard deviation $\sigma = 7$ What is the probability that a randomly chosen value will be greater than 18?
A) 0.9684
B) 0.6386
C) 0.9088
D) 0.0316

Question

Quizplus · Accepted Answer

To find the probability that a randomly chosen value from a normal distribution is greater than 18, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X = 18$, $\mu = 31$, and $\sigma = 7$. Calculating the Z-score gives us $Z = \frac{18 - 31}{7} = -1.857$. Looking up this Z-score in a standard normal distribution table or using a calculator, we find the probability of being less than 18 is approximately 0.0316. Therefore, the probability of being greater than 18 is $1 - 0.0316 = 0.9684$.

Use Technology to Solve the Following Problem: a Normal Population μ=31\mu = 31μ=31

Use Technology to Solve the Following Problem: a Normal Population $\mu = 31$