# 5. Normal Distributions and Standard (z) Scores Flashcards Preview

## Stats > 5. Normal Distributions and Standard (z) Scores > Flashcards

Flashcards in 5. Normal Distributions and Standard (z) Scores Deck (39)
1
Q

Normal Curve

A

A theoretical curve noted for its symmetrical bell-shaped form.

2
Q

z Score

A

A unit-free, standardized score that indicates how many standard deviations a score is above or below the mean of its distribution.

3
Q

How do you obtain a z score?

A

z = (X - µ) / σ

Express any origina score as a deviation from its mean (by subtracting its mean) and then split this deviation into standard deviation units (by dividing by its standard deviation).

4
Q

Express the following score as a z score:

Margaret’s IQ of 135, given a mean of 100 and a standard deviation of 15

A

-35/15 = 2.33

5
Q

Express the following score as a z score:

a score of 470 on the SAT math test, given a mean of 500 and a standard deviation of 100

A

-30/100 = -0.30

6
Q

Express the following score as a z score:

a daily production of 2100 loaves of bread by a bakery, given a mean of 2180 and a standard deviation of 50

A

80/50 = 1.60

7
Q

Express the following score as a z score:

Sam’s height of 69 inches, given a mean of 69 and a standard deviation of 3

A

0/3 = 0.00

8
Q

Express the following score as a z score:

a thermometer-reading error if -3 degrees, given a mean of 0 and a standard deviation of 2 degrees

A

-3/2 = -1.50

9
Q

Standard Normal Curve

A

The tabled normal curve for Z scores, with a mean of 0 and a standard deviation of 1.

10
Q

Give the 4 steps to find proportions for one score.

A
1. Sketch a normal curve and shade in the target area.
2. Plan your solution according to the normal table.
3. Convert X to z.
4. Find the target area.
11
Q

Assume that GRE scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion that corresponds to less than 400.

A

1) Sketch a normal curve and shade in the target area.
2) Plan solutions for the target areas (in terms of columns B, C, B’, or C’ of the standard normal table, as well as the fact that the proportion for either the entier upper half or lower half always equals .5000).
3) Converto to z scores and find the proportion that corresponds to the target area.

Column C’
z = (X - µ) / σ
z = (400 - 500) / 100 = -1
Proportion : .1587

12
Q

Assume that GRE scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion that corresponds to more than 650.

A

1) Sketch a normal curve and shade in the target area.
2) Plan solutions for the target areas (in terms of columns B, C, B’, or C’ of the standard normal table, as well as the fact that the proportion for either the entier upper half or lower half always equals .5000).
3) Converto to z scores and find the proportion that corresponds to the target area.

Column C
z = (X - µ) / σ
z = (650 - 500) / 100 = 1,5
Proportion : .0668

13
Q

Assume that GRE scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion that corresponds to less than 700.

A

1) Sketch a normal curve and shade in the target area.
2) Plan solutions for the target areas (in terms of columns B, C, B’, or C’ of the standard normal table, as well as the fact that the proportion for either the entier upper half or lower half always equals .5000).
3) Converto to z scores and find the proportion that corresponds to the target area.

Column B + 0.5000
z = (X - µ) / σ
z = (700 - 500) / 100 = 2
Proportion : 0.5000 + 0.4472 = 0.9772

14
Q

Assume that, when not interrupted artificially, the gestation periods for human fetuses approximate a normal curve with a mean of 270 days (9 months) and a standard deviation of 15 days. What proportion of gestation periods will be between 245 and 255 days?

A
1. Sketch a normal curve and shade in the targe area.
2. Plan your solution according to the normal table. The basic idea is to identify the target area with the difference between two overlapping areas whose values can be read from column C’.
Subtract the smaller area (<245) from the larger area (<255).
3. Convert X to z by expressing 255 (z = -1.00) and 245 (z = -1.67).
z = (X - µ) / σ
4. Find the target area.
- 1.00 => .1587
- 1.67 => .0475
5. 1587 - 0.0475 = 0.1112
15
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are more than 570.

A

C

z = 0.70

0.2420

16
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are less than 515

A

0.5000 + B

z = 0.15

0.5000 + 0.0596 = 0.5596

17
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are between 520 and 540

A

larger B - smaller B
or
larger C - smaller C

z = 0.20 ; z = 0.40

0.1554 - 0.0793 = 0.0761
or
0.4207 - 0.3446 = 0.0761

18
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are between 470 and 520

A

B’ + B

z = -0.30; z = 0.20

0.1179 + 0.0793 = 0.1972

19
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are more than 50 points above the mean

A

C

z = 0.50

0.3085

20
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are more than 100 points either above or below the mean

A
```C' + C
or 2(C)```

z = -1.00; z = 1.00

0.1587 + 0.1587 = 0.3174

21
Q

Assume that SAT math scores approximate a normal curve with a mean of 500 and a standard deviation of 100.

Find the proportion of scores that are within 50 points either above or below the mean.

A
```B' + B
or 2(B)```

z = -0.50; z = 0.50

0.1915 + 0.1915 = 0.3830

22
Q

Exam scores for a large psychology class approximate a normal curve with a mean of 230 and a standard deviation of 50. Furthermore, students are graded “on a curve”, with only the upper 20 percent being awarded grades of A. What is the lowest score on the exam that receives an A?

A
1. Sketch a normal curve and, on the correct side of the mean, draw a line representing the target score.
2. Plan your solution according to the normal table.
3. 3000 from column B (0.8000 - 0.5000)
4. 2000 from column C (1 - 0.8000)
5. Find z.
Scan column C to find 0.2000.
Or scan column B to find 0.3000.
In both cases, 0.84 is the closest.
6. Convert z to the target score [ X = µ + (z)(σ) ].
X = 230 + (0.84)(50) = 272

Therefore, 272 is the lowest score on the exam that receives an A.

23
Q

How to convert z score to original score (formula) ?

A

X = µ + (z)(σ)

24
Q

Assume that the annual rainfall in the San Francisco area approximates a normal curve with a mean of 22 inches and a standard deviation of 4 inches. what are the rainfalls for the more atypical years, defined as the driest 2.5 percet of all years and the wettest 2.5 percent of all years?

A
1. Sketch a normal curve. On either side of the mean, draw two lines representing the two target scores.
The smallest target score splits the total area into .0250 to the left and 0.9750 to the right, and the larger target score does the exact opposite.
2. Plan your solution according to the normal table.
The target z score of the smaller target score can be found by scanning either column B’ for .4750 or column C’ for .0250. The largest score is the opposite.
3. Find z.
4. 4750 corresponds to the z score -1.96 for the lowest score and 1.96 for the largest.
```4. Convert z to the target score.
X = µ + (z)(σ)
X = 22 + 1.96(4) = 22 + 7.84 = 29.84
and
X = 22 - 1.96(4) = 22 - 7.84 = 14.16 ```

2.5 % of the driest years registered less than 14.16 inches of rainfall, whereas 2.5 % of the wettest years registered more than 29.84 inches.

25
Q

Assume that the burning times of electric light bulbs approximate a normal curve with a mean of 1200 hours and a standard deviation of 120 hours.

If a large number of new lights are installed at the same time (possibly along a newly opened freeway), at what time will 1 % fail?

A

1200 + (-2.33)(120) = 920,4

26
Q

Assume that the burning times of electric light bulbs approximate a normal curve with a mean of 1200 hours and a standard deviation of 120 hours.

If a large number of new lights are installed at the same time (possibly along a newly opened freeway), at what time will 50 % fail?

A

1200 + (0.00)(120) = 1200

27
Q

Assume that the burning times of electric light bulbs approximate a normal curve with a mean of 1200 hours and a standard deviation of 120 hours.

If a large number of new lights are installed at the same time (possibly along a newly opened freeway), at what time will 95 % fail?

A

1200 + (1.65)(120) = 1398
or
1200 + (1.64)(120) = 1396.80

28
Q

Assume that the burning times of electric light bulbs approximate a normal curve with a mean of 1200 hours and a standard deviation of 120 hours.

If a new inspection procedure eliminates the weakest 8 percent of all lights before they are marketed, the manufacturer can safely offer customers a money-back guarantee on all lights that fail before ———- hours of burning time.

A

1200 + (-1.41)(120) = 1030.80

29
Q

Convert the following test score to z score.

Test score = 53
Mean = 50
Standard Deviation = 9

A

z = (X - µ) / σ

0.33

30
Q

Convert the following test score to z score.

Test score = 38
Mean = 40
Standard Deviation = 10

A

z = (X - µ) / σ

-0.20

31
Q

Convert the following test score to z score.

Test score = 45
Mean = 30
Standard Deviation = 20

A

z = (X - µ) / σ

0.75

32
Q

Convert the following test score to z score.

Test score = 28
Mean = 20
Standard Deviation = 20

A

z = (X - µ) / σ

0.4

33
Q

Standard Score

A

Any unit-free scores expressed relative to a known mean and a known standard deviation.

34
Q

Transformed Standard Score

A

A standard score that, unlike z score, usually lacks negative signs and decimal points.

35
Q

How to convert to transformed standard scores?

A

z’ = desired mean + (z) (desired standard deviation)

36
Q

Assume that each of the raw scores listed originates from a distribution with the specified mean and standard deviation. After converting each raw score into a z score, transform each z score into a series of new standard scores with means and standard deviations of 50 and 10, 100 and 15, and 500 and 100 respectively.

Raw Score = 24
Mean = 20
Standard Deviation = 5

A
```z = (X - µ) / σ
z = 4/5 = 0.80```

z’ = 50 + 0.8 (10) = 58

z’ = 100 + 0.8 (15) = 112

z’ = 500 + 0.8 (100) = 580

37
Q

Assume that each of the raw scores listed originates from a distribution with the specified mean and standard deviation. After converting each raw score into a z score, transform each z score into a series of new standard scores with means and standard deviations of 50 and 10, 100 and 15, and 500 and 100 respectively.

Raw Score = 37
Mean = 42
Standard Deviation = 3

A
```z = (X - µ) / σ
z = -5/3 = -1.67```

z’ = 50 - 1.67 (10) = 33.3

z’ = 100 - 1.67 (15) = 74.95

z’ = 500 - 1.67 (100) = 333

38
Q

Give the 4 steps to find proportions.

A

1) Sketch the normal curve and shade in the target area.
2) Plan the solution in terms of the normal table.
3) Convert X to z: z = (X - μ) / σ
4) Find the target area by entering either column A or A’ with z, and noting the corresponding proportion from column B, C, B’, or C’.

39
Q

Give the 4 steps to find scores.

A

1) Sketch the normal curve and, on the correct side of the mean, draw a line representing the target score.
2) Plan the solution in terms of the normal table.
3) Find z by locating the entry nearest to that desired in column B, C, B’, or C’ and reading out the correspond z score.
4) Convert z to the target score: X = μ + (z) (σ)