Question

unit 3 session 3 the arithmetic mean

1. a group of 20 students earned a class mean of 30 on a quiz. a second group of 30 students had a mean score of 45 on the same test. what is the mean score of the 50 students?

a. 32.5
b. 39.0
c. 41.0
d. 45.0

1. one strength of the mean as a measure of location is that it is

a. appropriate for nominal scale variables.
b. limited in further statistical analysis.
c. not affected by extreme scores.
d. useful for symmetrical distributions.

the table below gives the distribution of the ages of teachers in a school district.

age	number of teachers
40 - 44	36
35 - 39	77
30 - 34	47
25 - 29	15
total	200

1. what is the value of the mean age?

a. 35.2
b. 36.2
c. 37.2
d. 38.2

1. one weakness of the mean as a measure of central tendency is that it

a. cannot be used when data is complete.
b. is influenced largely by extreme scores.
c. is most appropriate for normal distributions.
d. uses few values in a distribution.

Explanation:

Question 2

Step1: Find total scores of each group

First group: 20 students, mean 30. Total = \(20 \times 30 = 600\)
Second group: 30 students, mean 45. Total = \(30 \times 45 = 1350\)

Step2: Find total scores of 50 students

Total scores = \(600 + 1350 = 1950\)

Step3: Calculate mean of 50 students

Mean = \(\frac{1950}{50} = 39.0\)

• Option A: Mean is not for nominal (categorical) data. Eliminate.
• Option B: Mean is used in many statistical analyses. Eliminate.
• Option C: Mean is affected by extreme scores (e.g., outliers). Eliminate.
• Option D: Mean is useful for symmetric distributions (e.g., normal), as it centers the data. Correct.

Step1: Find midpoint of each age group

• 45 - 49: Midpoint \(= \frac{45 + 49}{2} = 47\)
• 40 - 44: Midpoint \(= \frac{40 + 44}{2} = 42\)
• 35 - 39: Midpoint \(= \frac{35 + 39}{2} = 37\)
• 30 - 34: Midpoint \(= \frac{30 + 34}{2} = 32\)
• 25 - 29: Midpoint \(= \frac{25 + 29}{2} = 27\)

Step2: Calculate total age for each group (midpoint × number of teachers)

• 45 - 49: \(47 \times 25 = 1175\)
• 40 - 44: \(42 \times 36 = 1512\)
• 35 - 39: \(37 \times 77 = 2849\)
• 30 - 34: \(32 \times 47 = 1504\)
• 25 - 29: \(27 \times 15 = 405\)

Step3: Find total age of all teachers

Total age = \(1175 + 1512 + 2849 + 1504 + 405 = 7445\)

Step4: Calculate mean age

Mean age = \(\frac{7445}{200} = 37.225 \approx 37.2\)

Answer:

B. 39.0

Question 3