Normal Distributions

Finding Values

Section 5.4

z

From Areas to z-ScoresFind the z-score corresponding to a cumulative area of 0.9803.

–4 –3 –2 –1 0 1 2 3 4

0.9803

z

From Areas to z-Scores

Locate 0.9803 in the area portion of the table. Read the values at the beginning of the corresponding row and at the top of the

column. The z-score is 2.06.

Find the z-score corresponding to a cumulative area of 0.9803.

z = 2.06 correspondsroughly to the

98th percentile.

–4 –3 –2 –1 0 1 2 3 4

0.9803

Class Practice p234 1-11 odd

1.-2.053. 0.855. -0.167. 2.399.-1.64511. 0.84

Finding z-Scores from Areas

Find the z-score corresponding to the 90th percentile.

z0

.90

Finding z-Scores from Areas

Find the z-score corresponding to the 90th percentile.

z0

.90

The closest table area is .8997. The row heading is 1.2 and column heading is .08. This corresponds to z = 1.28.

A z-score of 1.28 corresponds to the 90th percentile.

Class Practice P234 13-23 odd

13. -2.32515.-0.2517. 1.17519. -0.67521. 0.67523. -0.385

Find the z-score with an area of .60 falling to its right.

.60

0 zz

Finding z-Scores from Areas

Find the z-score with an area of .60 falling to its right.

.60

.40

0 zzWith .60 to the right, cumulative area is .40. The closest area is .4013. The row heading is 0.2 and column heading is .05. The z-score is 0.25.

A z-score of 0.25 has an area of .60 to its right. It also corresponds to the 40th percentile

Finding z-Scores from Areas

Find the z-score such that 45% of the area under the curve falls between –z and z.

0 z–z

.45

Finding z-Scores from Areas

Find the z-score such that 45% of the area under the curve falls between –z and z.

0 z–z

The area remaining in the tails is .55. Half this area isin each tail, so since .55/2 = .275 is the cumulative area for the negative z value and .275 + .45 = .725 is the cumulative area for the positive z. The closest table area is .2743 and the z-score is 0.60. The positive z score is 0.60.

.45.275.275

Finding z-Scores from Areas

Class Practice P 234 27-33 odd

27. -1.645, 1.64529. -1.96, 1.9631. 0.32533. 1.28

From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard score of: (a) 2.33 (b) –1.75 (c) 0

To find the data value, x when given a standard score, z:

From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard score of: (a) 2.33 (b) –1.75 (c) 0

(a) x = 152 + (2.33)(7) = 168.31

(b) x = 152 + (–1.75)(7) = 139.75(c) x = 152 + (0)(7) = 152

To find the data value, x when given a standard score, z:

Finding Percentiles or Cut-off ValuesMonthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills?

z

Finding Percentiles or Cut-off ValuesMonthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills?

10%90%

Find the cumulative area in the table that is closest to 0.9000 (the 90th percentile.) The area 0.8997 corresponds to a z-score of 1.28.

x = 100 + 1.28(12) = 115.36.

$115.36 is the smallestvalue for the top 10%.

z

To find the corresponding x-value, use

Class Practice p 235 35 and 39

35. a. 68.52b. 62.14

39. a. 139.22b. 96.92

HW p 234-236 2-40 even