Question

A sample of n = 4 scores is selected from a normal population with μ = 30 and σ = 8. The probability of obtaining a sample mean greater than 34 is equal to the probability of obtaining a z-score greater than z = 2.00.

Answer 1

Answer:

False

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

$\mu = 30, \sigma = 8$

Sample of 4

This means that $n = 4, s = \frac{8}{\sqrt{4}} = 4$

Probability of obtaining a sample mean greater than 34:

This is 1 subtracted by the p-value of Z when X = 34. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{34 - 30}{4}$

$Z = 1$

Thus, the probability of obtaining a sample mean greater than 34 is equal to the probability of obtaining a z-score greater than z = 1.00, and the statement in this question is false.