In a survey, the guess at the mean age for the class had a mean of 21.9 and a standard deviation of 3.03.

Given that Normally distributed populations have a first quartile of Z=-0.67 and a third quartile of Z=.67, what does this suggest about whether or not these guesses are normally distributed? The actual first quartile is 20 and the third is 23.

I am so lost i don't know how to do the steps or if a Z scored formula should even be applied. Thank you

To determine whether the guesses at the mean age for the class are normally distributed, we can use the z-score formula and compare the z-scores for the first and third quartiles.

First, let's calculate the z-scores for the first and third quartiles using the formula:

Z = (X - μ) / σ

where:
- Z is the z-score
- X is the value (first quartile or third quartile)
- μ is the mean of the population (mean age guessed in the survey)
- σ is the standard deviation of the population (standard deviation of the guesses)

For the first quartile:
Z1 = (20 - 21.9) / 3.03

And for the third quartile:
Z3 = (23 - 21.9) / 3.03

Calculating these values gives us:
Z1 ≈ -0.63
Z3 ≈ 0.36

Now, let's compare these z-scores with the z-scores given in the problem statement. If the guesses are normally distributed, the z-scores for the first quartile and third quartile should match the z-scores (-0.67 and 0.67) for a normally distributed population.

Comparing the calculated z-scores with the given z-scores, we can see that Z1 ≈ -0.63 is close to -0.67, and Z3 ≈ 0.36 is significantly different from 0.67.

Based on this comparison, we can conclude that the guesses at the mean age for the class are not perfectly normally distributed. The deviation of Z3 suggests that there is some non-normality in the distribution of the guesses. However, without more information or a larger sample, it is difficult to determine the exact nature of the distribution.

Please note that this analysis assumes that the sample used to make the guesses is representative of the class and that the population of guesses follows a normal distribution.