Question

a. for a random sample of $n = 4$ people?
b. for a random sample of $n = 16$ people?
c. for a random sample of $n = 25$ people?

1. a normal distribution has a mean of $mu = 58$ and a standard deviation of $sigma = 12$.

a. what is the probability of randomly selecting a score less than $x = 52$?
b. what is the probability of selecting a sample of $n = 9$ scores with a mean less than $m = 52$?

Explanation:

Step1: Calculate the z - score for part a

The formula for the z - score is \(z=\frac{X-\mu}{\sigma}\). Given \(\mu = 58\), \(\sigma=12\), and \(X = 52\).

$$z=\frac{52 - 58}{12}=\frac{-6}{12}=- 0.5$$

Using the standard normal distribution table, the probability \(P(Z\lt - 0.5)\) is \(0.3085\).

Step2: Calculate the standard error for part b

The formula for the standard error is \(\sigma_{M}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 12\) and \(n = 9\).
\(\sigma_{M}=\frac{12}{\sqrt{9}}=\frac{12}{3}=4\)

Step3: Calculate the z - score for part b

The formula for the z - score for a sample mean is \(z=\frac{M-\mu}{\sigma_{M}}\). Given \(\mu = 58\), \(M = 52\), and \(\sigma_{M}=4\).

$$z=\frac{52 - 58}{4}=\frac{-6}{4}=-1.5$$

Using the standard normal distribution table, the probability \(P(Z\lt - 1.5)\) is \(0.0668\).

Answer:

a. The probability of randomly selecting a score less than \(X = 52\) is \(0.3085\).
b. The probability of selecting a sample of \(n = 9\) scores with a mean less than \(M = 52\) is \(0.0668\).