Question 1

Please give me an EXAMPLE that illustrates why it is important to know the standard deviation of a distribution as well as the mean.

Accepted Answer

Knowing that the average amount of rainfall in California over the last 10 years is 40 inches per year may be misleading. Do we get about 40 inches of rain in CA every year (small s.d.)? Or do we get 10 inches some years and 100 inches others (large s.d.). If your health or business depend on reliable amounts of rain (e.g., farmers), you'd want to know the s.d. in addition to the mean.

Question 2

Suppose I want to know how many household pets Americans have owned in their lifetimes before they turn 30 years old. I select a sample of 3 Americans, each of whom is 30 years old, and ask them how many pets they have owned in their lifetimes. The first person I ask has owned 6 animals, the second person has owned 2, and the third person has owned 11. Please answer the following questions based on these data.
a. Calculate and report the standard deviation for this distribution, then wrap words around it. What does it tell you?
b. When calculating the standard deviation, you had to decide whether to use N or n - 1 in the denominator. Which did you use, and why?
c. When calculating the standard deviation, you had to produce a sum of the squared deviations. Please report that number, tell me how you calculated it, and tell me what it means.
d. Calculate and report the range for this distribution. Please show me how you calculated the range.

Accepted Answer

a. (2 - 6.33)² = 18.75 (6 - 6.33)² = .11 (11 - 6.33)² = 21.81 SS = 18.75 + .11 + 21.81 = 40.67 40.67/2 = 20.34 = s² √20.34 = 4.51 = s So the average deviation, or difference, between scores in the sample and the mean for the sample is 4.51 pets owned by the age of 30. b.Used n - 1 because this is a sample, not a population. This was stated in the research description. When using a sample to estimate the population standard deviation, you must adjust the denominator by using n - 1. c. SS = 40.67. This means that the sum of the squared deviations (between each individual score in the distribution and the sample mean) was 40.67. I calculated it by subtracting the mean for the sample from each score, squaring the difference, and adding up the squared deviations. d. 11 - 2 = 9

Question 3

Suppose that my 3 siblings and I are the following heights, in inches.
72 69 77 66
a. Calculate and report the range of this distribution.
b. Calculate and report the variance of this distribution.
c. Tell me how you calculated this.
d. Calculate and report the standard deviation of this distribution.
e. Write a sentence or two in which you tell me, using your own words, what a standard deviation is (in general, not this particular standard deviation).
f. Wrap words around the mean and standard deviation of this distribution. What do they tell you about the distribution of scores?
g. Pretend that this distribution of scores represents a sample rather than a population. Calculate and report the standard deviation.
h. Explain why the standard deviation you reported for question 1d differs from the standard deviation you reported for question 1g.
i. Suppose that in the population of birds that live in the trees on a university campus, the average number of feathers is 1000 with a standard deviation of 200. Wrap words around that standard deviation. What does it tell you, exactly?

Accepted Answer

a. 77 - 66 = 11 b. 16.5 c. I took each score, subtracted the mean from it, and squared it: (72-71)² = 1, (69-71)² = 4, (77-71)² = 36, (66-71)² = 25 Then you add these all together: 1+4+36+25 = 66 Then you divide this sum by the number of cases in the population: 66/4 = 16.5 d. The square root of the variance is the standard deviation. So, sq. root of 16.5 = 4.06. e. It is the average difference between the mean and the scores in the distribution. f. The average height of my siblings and I is 71 inches with a standard deviation of 4.06 inches, so the average difference between out individual heights and the mean height is 4.06 inches. g.This is a population because my siblings and I are all of the members of a group (my parents' children). So when calculating the s.d., I divided the variance by N, which was 4. But if this had been a sample, I would have divided by n - 1, which is 3. So 66/3 = 22, and the square root of 22 is 4.69. So had this been a sample, the s.d. would have been 4.69. h. Because for a sample we us n - 1 in the denominator, not N. i. The average difference between the mean of the population and the individual scores in the population is 200 birds.

Exam 3: Measures of Variability

Please give me an EXAMPLE that illustrates why it is important to know the standard deviation of a distribution as well as the mean.