Section 5.3
Normal Distributions
Finding Probabilities
Probabilities and Normal Distributions
If a random variable, x is normally distributed, the probability that
x will fall within an interval is equal to the area under the curve in
the interval.
IQ scores are normally distributed with a mean of 100 and a
standard deviation of 15. Find the probability that a person
selected at random will have an IQ score less than 115.
100
115
To find the area in this interval, first find the standard score equivalent to x = 115.
Probabilities and Normal Distributions
Normal Distribution
Find P(x < 115).
100 115
SAME
SAME
Standard Normal Distribution
Find P(z < 1).
0 1
P(z < 1) = 0.8413, so P(x <115) = 0.8413
Application
Monthly utility bills in a certain city are normally distributed with
a mean of $100 and a standard deviation of $12. A utility bill is
randomly selected. Find the probability it is between $80 and
$115.
Normal Distribution
P(80 < x < 115)
P(–1.67 < z < 1.25)
0.8944 – 0.0475 = 0.8469
The probability a utility bill is between
$80 and $115 is 0.8469.
From Areas to z-Scores
Find the z-score corresponding to a cumulative area of 0.9803.
z = 2.06 corresponds
roughly to the
98th percentile.
0.9803
–4
–3
–2
–1 0
1
2
3
4
z
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.
Finding z-Scores from Areas
Find the z-score corresponding to the 90th percentile.
.90
0
z
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.
Finding z-Scores from Areas
Find the z-score with an area of .60 falling to its right.
.40
.60
z
0
z
With .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 zscore 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.
.275
.275
.45
–z
0
z
The area remaining in the tails is .55. Half this area is
in 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.
From z-Scores to Raw Scores
To find the data value, x when given a standard score, z:
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
Finding Percentiles or Cut-off Values
Monthly 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?
$115.36 is the smallest
value for the top 10%.
90%
10%
z
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.
To find the corresponding x-value, use
x = 100 + 1.28(12) = 115.36.
Homework : 1-37 (odd) pgs. 242-243
42-46 even pgs.243-244. Day 2:Homework : 2-36
(even) pgs. 242-243
41-45 odd pgs.243-244.