Question

QUESTION 1

The normal curve is particularly useful as a model for

	a.	data in which mean and median differ
	b.	many populations of psychological and educational data
	c.	distributions of sample statistics
	d.	both (b) and (c) above

QUESTION 2

A distribution has a mean of 60 and a standard deviation of 8. For a score of 72, the equivalent z score

	a.	is +1.5
	b.	is between 0 and +1.0
	c.	is + 1.2
	d.	cannot be determined without further information

QUESTION 3

A z score in a given distribution is -.5. If the mean of this distribution is 130 and the standard deviation is 20, the equivalent raw score in that distribution is

	a.	cannot be determined without further information
	b.	129.5
	c.	120
	d.	110

QUESTION 4

College Board scores have a mean of 500 and a standard deviation of 100. On this scale, a score of 450 is the equivalent of

	a.	z = -.50
	b.	z = -1.0
	c.	z = 0
	d.	Not enough information to determine

QUESTION 5

If a set of raw scores is positively skewed, the set of z scores derived from them will be

	a.	symmetrical, but not normal
	b.	positively skewed
	c.	very close to a normal distribution, but slightly positively skewed.
	d.	normally distributed

QUESTION 6

The percent of cases in a normal distribution falling between z = -.67 and a z = +.67 is approximately

	a.	.25
	b.	1.34
	c.	.67
	d.	.50

QUESTION 7

In a normal distribution, what proportion of cases falls below z = +1.40?

	a.	.0808
	b.	.9192
	c.	.4192
	d.	.5808

QUESTION 8

In a normal distribution of 200 cases, how many fall between z = -1.5 and z = +1.5?

	a.	173
	b.	43
	c.	27
	d.	87

QUESTION 9

In a normal distribution of scores with mean = 50 and standard deviation = 10, what proportion of cases falls above a score of 62?

	a.	.0478
	b.	.3849
	c.	.1151
	d.	.7698

QUESTION 10

In a normal distribution with mean = 50 and standard deviaation = 10, the bottom 30% of the cases fall below what score (rounded)

	a.	40
	b.	45
	c.	47
	d.	38

Answer 1

1.
d) both b and c

2.
Mean = 60
SD = 8
X =72
z = (X - Mean) / SD
z = (72 - 60) / 8 = 1.5
a) +1.5

3.

z = -0.5
Mean = 130
SD = 20
z = (X - Mean) / SD
X = z.SD + Mean
X = (-0.5 x 20) + 130
X = 120
c) 120

4.

Mean = 500
SD = 100
X = 450
z = (X - Mean) / SD
z = (450 - 500) / 100 = -0.5
a) -0.50

5.
b) positively skewed

6.

Percent of cases b/w z = -0.67 and z = 0 = Percent of cases b/w z = 0 and z = 0.67
From standard normal table z = 0.67 => 0.7486
B/w z = 0.67 and z = 0 => 0.7486 - 0.50 = 0.2486
The percent of cases in a normal distribution falling between z = -.67 and a z = +.67 = 0.2486 x 2 = 0.4972 ~ 0.50
d) 0.50

7.
From standard normal table ,
In a normal distribution, what proportion of cases falls below z = +1.40 = .9192
b) 0.9192

8.
Proportion of cases falling below +1.5 = 0.9332
Proportion b/w 0 and 1.5 = 0.9332 - 0.5000 = 0.4332
Proportion b/w -1.5 and +1.5 = 0.4332 x 2 = 0.8664
Out of 200 cases, number of cases falling b/w -1.5 and +1.5 = 0.8664 x 200 = 173.28
a) 173

9.

Mean = 50
SD = 10
X = 62
z = ( X - Mean) / SD
z = (62-50)/10 = 1.2
P(Z< 1.2) from standard normal table = 0.8849

10.
z-score corresponding to bottom 30% - look for value closest to 0.30 in standard normal table i.e 0.3015
z = -0.52
Mean = 50
SD = 10
z = (X- Mean) / SD
X = (-0.52 x 10) + 50 = 50 - 5.2 = 44.8
b) 45