Really struggling with this stats problem—any help would mean a lot

Question

The 5-factor model is a popular theory of personality, the 5 factors being Openness, Conscientiousness, Extraversion, Agreeableness, and Neuroticism. The Agreeableness scores of a random sample of 200 Buddhist monks in Thailand have a mound-shaped, symmetrical distribution and the mode is 100. One monk, whose score was 130, was found to fall at P97.5.

a) Approximately how many of the monks in this sample would obtain Agreeableness scores between 90 and 120?

b) What percentage of the sample would be expected to score higher than 1.38 standard deviations below the mean?

Answer 1

This question wants you to assume mound means "bell-shaped" in this case. The idea is that because the distribution is mound/bell-shaped with no significant outliers and the sample size(200) is very large, that you can assume the distribution is approximately normal. With normal distributions you can apply the Empirical Rule, which states that all area under a bell curve within 1, 2, 3 standard deviations away from the mean(center of the bell curve) consists of roughly 68%, 95%, and 99.7% of the data respectively.

One monk is noted to have a score of 130 at the 97.5% mark. We want to use this information and what we know about the Empirical Rule to come to the conclusion that if 95% of the data lies within 2 standard deviations away from the center of the bell curve, then 5% of the data is left over. Since there are two tails in a normal distribution, 2.5% of the data lies on the left tail and thus the 2.5% of the data on the left tail plus the 95% of the data that lies within 2 standard deviations from the center adds up to 97.5% total data. Thus, the monk with 130 lies exactly 2 standard deviations above the mean(100), and by performing the calculation (130 - 100)/2, we realize the standard deviation of the distribution is 15.

Our mean is 100 and our standard deviation is 15, so now for part a) we want to calculate the z-scores for 90 and 120. Using the formula for calculating z-scores, (value - mean)/standard deviation, we see that the z-scores for 90 and 120 are -.67 and 1.33 respectively. Using a z-score table(you can look for one with both positive and negative z-scores online), we see that the -.67 z-score gives us a value of .25143 and the 1.33 z-score gives us a value of .90824. Performing the calculation of .90824 - .25143 tells us the proportion of monks between 90 and 120 is .65681. The reason we do subtraction is that the .90824 tells us the proportion of monks below 120, while subtracting .25143, the proportion associated with 90, eliminates the monks below 90. Multiplying .65681 by 200, our initial sample of monks, and rounding to the nearest whole number gives us 131. So our answer to part a) is 131 monks.

For part b), we are looking for the percentage of monks higher than 1.38 standard deviations below the mean. The statement "below the mean" indicates that our z-score is not +1.38 but -1.38, so we want to use the z-table to look for -1.38 and we see from this that our proportion is .08379. We want to do the calculation 1 - .08379 because we are looking for monks higher than -1.38 standard deviations from the mean which is the opposite of lower in this case. The proportions on the z-table regard being below various z-scores. 1 - .08379 = .91621, and converting .91621 to a percentage tells us that approximately 91.621% of monks are expected to score higher than 1.38 standard deviations below the mean. Thus, 91.621% is our answer for part b).

Answer 2

This is a bad problem. The idea is that you are supposed to show that you can calculate probabilities with a normal distribution. In practice, you should not assume that an "mound shaped, symmetrical distribution" is a normal distribution. However, perhaps in your course your instructor is telling you to make that assumption, and then probably to report estimates to several unwarranted decimal places to make it easier to check if you are executing the techniques covered. These are not good habits to use outside of the course.

I assume you have a z-table. If not, you can look one up.

A normal distribution can be determined by the mean and standard deviation. You are given that the mean is 100 but you need to determine the standard deviation from the information that 130 is at the 97.5th percentile. The 97.5th percentile for a normal distribution is at a z-score of 1.96. (Do an inverse look-up of a probability 0.9750 in a z-table to get a z-score of 1.96.) This is worth memorizing because this specific value comes up a lot. A score that is 130-100=30 points above the mean is 1.96 standard deviations above the mean, so the standard deviation is 30/1.96 = 15.31. It could be that your course is using 2 in place of 1.96, which would give you an estimate of 15 for the standard deviation. Use this to translate scores of 120 and 90 to z-scores using the formula (x-mean)/standard deviation, then take the difference between the probabilities for those z-scores in your z-table to get the probability of a score between 90 and 120.

The z-table's value for a z=1.38 is the probability the z-score is less than 1.38. You need the complement, one minus that probability, the probability the z-score is greater than 1.38. Multiply the complement by 100 to convert the probability to a percentage.