Share
Explore BrainMass

# Normal distribution, z score, survey, sampling methods

14. On a standard measure of hearing ability, the mean is 300 and the standard deviation is 20. Give the Z scores for persons who score (a) 340, (b) 310, and (c) 260. Give the raw scores for persons whose Z scores on this test are (d) 2.4, (e) 1.5, (f) 0, and (g) _4.5.

15. A person scores 81 on a test of verbal ability and 6.4 on a test of quantitative ability. For the verbal ability test, the mean for people in general is 50 and the standard deviation is 20. For the quantitative ability test, the mean for people in general is 0 and the standard deviation is 5. Which is this person's stronger ability: verbal or quantitative? Explain your answer to a person who has never had a course in statistics.

22. Suppose you want to conduct a survey of the attitude of psychology graduate students studying clinical psychology towards psychoanalytic methods of psychotherapy. One approach would be to contact every psychology graduate student you know and ask them to fill out a questionnaire about it. (a) What kind of sampling method is this? (b) What is a major limitation of this kind of approach?

25. You are conducting a survey at a college with 800 students, 50 faculty members, and 150 administrative staff members. Each of these 1,000 individuals has a single listing in the campus phone directory. Suppose you were to cut up the directory and pull out one listing at random to contact. What is the probability it would be (a) a student, (b) a faculty member, (c) an administrative staff member, (d) a faculty or administrative staff member, and (e) anyone except an administrative staff member? (f) Explain your answers to someone who has never had a course in statistics.

#### Solution Preview

14.  On a standard measure of hearing ability, the mean is 300 and the standard deviation
is 20. Give the Z scores for persons who score (a) 340, (b) 310, and (c)
260. Give the raw scores for persons whose Z scores on this test are (d) 2.4, (e)
1.5, (f) 0, and (g) _4.5.

a) Mean=&#956;= 300
Standard deviation =&#963;= 20
x= 340
z=(x-&#956;)/&#963;= 2 =(340-300)/20

b) Mean=&#956;= 300
Standard deviation =&#963;= 20
x= 310
z=(x-&#956;)/&#963;= 0.5 =(310-300)/20

c) Mean=&#956;= 300
Standard deviation =&#963;= 20
x= 260
z=(x-&#956;)/&#963;= -2 =(260-300)/20

d) Mean=&#956;= 300
Standard deviation =&#963;= 20
z=(x-&#956;)/&#963;= 2.4
x=&#956;+z&#963;= 348 =300+(2.4*20)

e) Mean=&#956;= 300
Standard deviation =&#963;= 20
z=(x-&#956;)/&#963;= 1.5
x=&#956;+z&#963;= 330 =300+(1.5*20)

f) Mean=&#956;= 300
Standard deviation =&#963;= 20
z=(x-&#956;)/&#963;= 0
x=&#956;+z&#963;= 300 =300+(0*20)

g) Mean=&#956;= 300
Standard deviation ...

#### Solution Summary

Answers to questions on Normal distribution, z score, survey, sampling methods

\$2.19