Questions: The scores of students on the SAT college entrance examinations at a certain high school had a normal distribution with mean μ=551.6 and standard deviation σ=27.1. (a) What is the probability that a single student randomly chosen from all those taking the test scores 556 or higher? ANSWER: For parts (b) through (d), consider a simple random sample (SRS) of 30 students who took the test. (b) What are the mean and standard deviation of the sample mean score x̄, of 30 students? The mean of the sampling distribution for x̄ is: The standard deviation of the sampling distribution for x̄ is: (c) What z-score corresponds to the mean score x̄ of 556? ANSWER: (d) What is the probability that the mean score x̄ of these students is 556 or higher?

Solution Steps

To find the probability that a single student scores 556 or higher, we calculate the Z-score corresponding to \( X = 556 \):

\[ Z = \frac{X - \mu}{\sigma} = \frac{556 - 551.6}{27.1} \approx 0.1624 \]

Using the standard normal distribution, we find:

\[ P(X \geq 556) = 1 - P(Z < 0.1624) = \Phi(\infty) - \Phi(0.1624) \approx 0.4355 \]

Thus, the probability that a single student scores 556 or higher is:

\[ \boxed{P \approx 0.4355} \]

For a simple random sample of \( n = 30 \) students, the mean and standard deviation of the sample mean \( \bar{x} \) are given by:

\[ \text{Mean of } \bar{x} = \mu = 551.6 \]

\[ \text{Standard deviation of } \bar{x} = \frac{\sigma}{\sqrt{n}} = \frac{27.1}{\sqrt{30}} \approx 4.9478 \]

Thus, the mean and standard deviation of the sample mean are:

\[ \text{Mean: } \boxed{551.6} \] \[ \text{Standard Deviation: } \boxed{4.9478} \]

To find the Z-score corresponding to the sample mean \( \bar{x} = 556 \):

\[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{556 - 551.6}{4.9478} \approx 0.8893 \]

Thus, the Z-score for the sample mean score of 556 is:

\[ \boxed{0.8893} \]

Final Answer