Question

Presented below are the means, standard deviations, and scores on various interval-ratio variables that are assumed to be normally distributed. For each problem, either use the Z table to find what percentage of the area under the normal curve lies above or below a given score, or compare two scores as requested. Show your work and state your answers in sentence form.

6. The average number of cats per U.S. household is 2.1, with a standard deviation of .9. Assume that the number of cats you own is a measure of coolness: if my partner and I have four cats, what percentage of U.S. households can be considered lamer than ours (i.e., what percentage of households have fewer cats)? [4 points]

7. Data from the 2018 European Social Survey (ESS)indicates that, among European households, the average number of inhabitants is 2.62 with a standard deviation of 1.38. Assuming data collected from the ESS are representative of the entire European population, what percentage of households throughout Europe have morethan five inhabitants?  [4 points]

8. The average number of Facebook (FB) friends among U.S. users is 338, with a standard deviation of 150. If Ryan only has 170 FB friends, what percentage of American FB users have more friends than him?  [4 points]

9. In Fall 2017, the average final grade in Ryan’s SOC 2830 course was 80%, with a standard deviation of 11%. The following semester, in Spring 2018, the average final grade was 79% (s = 8%). During which semester did a student earning 90% in the course perform better relative to their classmates? [5 points]

Answer 1

Since the data are normally distributed, we will compute the Z-score.

6. Z= (X- mean)/ SD= (4-2.1)/0.9= 1.9/0.9= 2.1111111111

Hence P(X<4)= P(Z<2.1111111111)= 0.9826 (using standard normal table).

Thus, we see that the average no. of cats per household being less than 4 covers an area of 0.9826( total area being 1), hence we conclude that 98.26% of U.S. households can be considered lamer than the one having 4 cats.

7. Z= (X- mean)/SD= (5-2.62)/1.38= 2.38/1.38= 1.7246376812

Hence P(X>5)= P(Z>1.7246376812)= 1- P(Z<1.7246376812)= 1- 0.9573= 0.0427

Thus, we see that the average no. of inhabitants, among European households, being greater than 5 covers an area of 0.0427, hence we can conclude that 4.27% of European households have more than 5 inhabitants.

8. Z= (X- mean)/SD= (170-338)/150= -168/150= -1.12

Hence, P(X>170)= P(Z>-1.12)= P(Z<1.12)= 0.8686

Thus, we see that the average no. of FB friends, among US users, being more than 170 covers an area of 0.8686, hence we can conclude that 86.86% of US users have more than 170 FB friends.