A normal population has a mean $\mu = 33$ and standard deviation $\sigma = 9$. What is the probability that a randomly chosen value will be greater than 44 ?
A) 0.1112
B) 0.8554
C) 0.6915
D) 0.8888

Question

Quizplus · Accepted Answer

To find the probability that a randomly chosen value from a normal distribution is greater than a specific value, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. For $X = 44$, $\mu = 33$, and $\sigma = 9$, the Z-score is $Z = \frac{44 - 33}{9} = \frac{11}{9} \approx 1.22$. Looking up this Z-score in a standard normal distribution table or using a calculator, we find the probability of being less than 44 is approximately 0.8888. Therefore, the probability of being greater than 44 is $1 - 0.8888 = 0.1112$.

A Normal Population Has a Mean μ=33\mu = 33μ=33 And Standard Deviation

A Normal Population Has a Mean $\mu = 33$ And Standard Deviation