In the population of male students, motivation scores follow a normal distribution with an average of 32.4 and a standard deviation of 7. A random sample of 16 male students is to be taken. They are all asked to answer the motivation questionnaire so that their motivation scores can be determined. What is the probability that the sample mean score will exceed 35?
A)

Question

In the population of male students, motivation scores follow a normal distribution with an average of 32.4 and a standard deviation of 7. A random sample of 16 male students is to be taken. They are all asked to answer the motivation questionnaire so that their motivation scores can be determined. What is the probability that the sample mean score will exceed 35?
A) < 0.0001
B) 0.0681
C) 0.1977
D) 0.3557

Quizplus · Accepted Answer

To find the probability that the sample mean score will exceed 35, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 7$ and $n = 16$, so $SE = \frac{7}{\sqrt{16}} = \frac{7}{4} = 1.75$.Next, we calculate the z-score to find how many standard errors the score of 35 is away from the mean. The z-score formula is $z = \frac{\bar{x} - \mu}{SE}$, where $\bar{x}$ is the sample mean, and $\mu$ is the population mean. Substituting the given values, we get $z = \frac{35 - 32.4}{1.75} = \frac{2.6}{1.75} \approx 1.4857$.Finally, we look up the z-score of 1.4857 in the standard normal distribution table or use a calculator to find the probability to the left of this z-score, which is approximately 0.9319. The question asks for the probability that the sample mean score will exceed 35, so we subtract this value from 1 to get the probability to the right of the z-score: $1 - 0.9319 = 0.0681$.

In the Population of Male Students, Motivation Scores Follow a Normal