TOPIC 4
APPLICATIONS OF THE NORMAL DISTRIBUTION
The Standard Normal Distribution
-3
-2
-1
0
1
2
3
Z
1.62
The shaded area between the middle (0) and any value of Z is given in the tables.
•
•
Total area in any distribution is 1.
Normal Distribution is symmetric.
Examples – Using Tables to find the Area between Z=0 and the Following Z
values: Z
0.4
1.0
1.47
-1.47
3.82
-4.69
AREA
.1554
.3413
.4292
.4292
.5000
.5000
Using the Standard Normal Distribution for Real Situations
Example : The heights of adult males is normally distributed with mean 170cm and
Standard Deviation 10cm.
Q1. Find the probability of a male between 180 and 190 cm.
Method:
a) Draw diagram and shade area required.
b) Use formula : Z =
X −μ
σ
To calculate Z values of the boundaries of shaded area
Z 180 =
180 − 170
=1
10
190 − 170
=2
10
c) Find areas in tables
Z
Area
1
.3413
2
.4772
Z 190 =
d) Add or subtract table areas to get probability required.
P(180 ≤ x ≤ 190) = P(1 ≤ Z ≤ 2)
= .4772 - .3413
= .1359
Further Examples
Q2. Taller than 190cm
Tables : .4772
Z=
190 − 170
=2
10
P(X>190) = .5 - .4772
= .0228
Q3. Shorter than 180cm
Z=
180 − 170
=1
10
Tables: .3413
P(X<180) = .5 + .3413
= .8413
Q4. Shorter than 165cm.
Tables : .1915
Z=
165 − 170
= −0.5
10
P(X<165) = .5000 - .1915
= .3085
Q5. Between 165 and 177cm.
165 − 170
= −0.5
10
177 − 170
=
= 0.7
10
Z 165 =
Tables : .1915
Z 177
Tables : .2580
P(165<X<177) = .1915 + .2580
= .4495
Using the Procedure in Reverse
Example - How high would a ceiling have to be so that 99% of males could stand
up straight?
a) Find suitable tables area.
b) Look up Z (Nearest).
Z = 2.33
c) Use formula
Z=
X −μ
2.33 =
σ
X − 170
10
X = 193.3 cm.
To Solve for X
Z=
X −μ
σ
X = Zσ + μ
Are Different
Versions of the
same Formula
μ = X − Zσ
σ =
X −μ
Z
Distribution of Sample Means
Example : Take a sample of 4 people from a population with mean 170 cm and
Standard Deviation 10 cm. What is the probability that the average of this sample is
greater than 180 cm?
μ = 170
σ = 10
n=4
ZX =
X −μ
σ
=
n
180 − 170
10
4
ZX = 2
Tables : .4772
P ( X > 180) = .5000 − .4772
= .0228
Example : What is the probability that the mean of a sample of 25 will be between
160 and 175 cm if the population mean is 170 cm and Standard Deviation is 10 cm?
Z 160 =
160 − 170
= −5
10
Tables : .5000
25
Z 175 =
175 − 170
= 2.5
10
Tables : ..4938
25
P(160 < X < 175) = .5000 + .4938 = .9938
Estimation of Population Mean from Sample Mean
Example : A sample of 36 people has an average weight of 80 Kg. What is the
average weight of the population?
Simple Answer : µ = 80 Kg. (Best Guess)
If we know that the Standard Deviation of the population is 15 Kg, we can find the
95% confidence limits for the population average using the formula.
μ = X ±Z
σ
n
μ = 80 ± 1.96
(15)
36
= 80 ± 4.9
or
between 75.1 and 84.9.
Example : Find the population mean with 99% confidence if a sample of 100 people
has a mean of 80 Kg and the population Standard Deviation is 10 Kg.
μ = 80 ± 2.58
(10)
100
= 80 ± 2.58
OR
77.42 ↔ 82.58
Finding the Required Sample Size to Estimate a Population Mean
Example : How large a sample would be needed to estimate a population mean
income to an accuracy of $10 with 95% confidence if the Standard Deviation is about
$150?
⎛ Zσ ⎞
n=⎜
⎟
⎝ w ⎠
2
⎛ (1.96)(150) ⎞
=⎜
⎟
10
⎝
⎠
= 864.36 ≈ 865
2
Example : Find the sample size needed to estimate the population mean weight to an
accuracy of 3 Kg with 90% confidence if the heaviest person is about 130 Kg. and the
lightest person is about 70 Kg.
Range 130 − 70
=
= 15
4
4
w=3
σ ≈
⎛ (1.645)(15) ⎞
n=⎜
⎟
3
⎝
⎠
= 67.65 ≈ 68
2