Download The Standard Normal Distribution

The Standard
Normal Distribution
1
The Standard Score
The standard score, or z-score, represents the number of
standard deviations a random variable x falls from the
mean.
2
The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
–4 –3 –2 –1
0 1
3
2 3
4
z
Cumulative Areas
The
total
area
under
the curve
is one.
–3 –2 –1 0 1 2 3
z
• The cumulative area is close to 0 for z-scores close
to –3.49.
• The cumulative area for z = 0 is 0.5000.
• The cumulative area is close to 1 for z-scores close to
4
3.49.
Cumulative Areas
Find the cumulative area for a z-score of –1.25.
0.1056
–3 –2 –1 0 1 2 3
z
Read down the z column on the left to z = –1.25 and across to
the column under .05. The value in the cell is 0.1056, the
cumulative area.
The probability that z is at most –1.25 is 0.1056.
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Finding Probabilities
To find the probability that z is greater than a given
value, subtract the cumulative area in the table
from 1.
Find P(z > –1.24).
0.1075
0.8925
z
–3 –2 –1 0 1 2 3
The cumulative area (area to the left) is 0.1075. So the area
to the right is 1 – 0.1075 = 0.8925.
P(z > –1.24) = 0.8925
6
Finding Probabilities
To find the probability z is between two given values, find the
cumulative areas for each and subtract the smaller area from
the larger.
Find P(–1.25 < z < 1.17).
–3 –2 –1 0 1 2
3
z
2. P(z < –1.25) = 0.1056
1. P(z < 1.17) = 0.8790
3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734
7
Summary
To find the probability that z is less
than a given value, read the
corresponding cumulative area.
-3 -2 -1 0 1 2 3
z
To find the probability is greater
than a given value, subtract the
cumulative area in the table from 1.
-3 -2 -1 0 1 2 3
z
To find the probability z is
between two given values, find the
cumulative areas for each and
subtract the smaller area from the
8
larger.
-3 -2 -1 0 1 2 3
z
Normal Distributions
Finding Probabilities
9
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.
10