Transcript Holt McDougal Algebra 2

Significance of Experimental Results
• How do we use tables to estimate
areas under normal curves?
•How do we recognize data sets that
are not normal?
Holt McDougal Algebra 2
Significance of Experimental Results
Example 1: Joint and Marginal Relative Frequencies
Jamie can drive her car an average of 432 gallons per tank of gas, with a
standard deviation of 36 miles. Use the graph to estimate the probability that
Jamie will be able to drive more than 450 miles on her next tank of gas.
The area under the normal
curve is always equal to 1.
Each square on the grid has
an area of 10(0.001) = 0.01.
Count the number of grid
squares under the curve for
values of x greater than 450.
There are about 31 squares under the graph, so the probability is about 31(0.01)
= 0.31 that she will be able to drive more than 450 miles on her next tank of gas.
Holt McDougal Algebra 2
Significance of Experimental Results
Example 2: Joint and Marginal Relative Frequencies
Estimate the probability that Jamie will be able to drive less than 400
miles on her next tank of gas?
There are about 19
squares under curve
less than 400, so the
probability is about
19(0.01) = 0.19 that
she will be able to
drive less than 400
miles on the next tank
of gas.
Holt McDougal Algebra 2
Significance of Experimental Results
Holt McDougal Algebra 2
Significance of Experimental Results
Example 3: Using Standard Normal Values
Scores on a test are normally distributed with a mean of 160 and
a standard deviation of 12.
a. Estimate the probability that a randomly selected student
scored less than 148.
z
x

148  160

 1
12
Px  148  0.16
The probability of scoring less than 148 is 0.16.
Holt McDougal Algebra 2
Significance of Experimental Results
Example 3: Using Standard Normal Values
Scores on a test are normally distributed with a mean of 160 and
a standard deviation of 12.
b. Estimate the probability that a randomly selected student
scored between 154 and 184.
z
x

154  160

 0.5
12
184  160

2
12
Px  184  0.98
Px  154  0.31
P154  x  184  0.98  0.31  0.67
The probability of scoring between 154 and 184 is 0.67.
Holt McDougal Algebra 2
Significance of Experimental Results
Example 4: Using Standard Normal Values
Scores on a test are normally distributed with a mean of 142 and
a standard deviation of 18. Estimate the probability of scoring
above 106.
z
x

106  142

 2
18
For greater than,
use the opposite sign
Px  106  0.98
The probability of scoring above 106 is 0.98.
Holt McDougal Algebra 2
Significance of Experimental Results
Scores on a test are normally distributed with a mean of 80
and a standard deviation of 5. Use the table below to find
each probability.
z
2.5
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
Area
0.01
0.02
0.07
0.16
0.31
0.5
0.69
0.84
0.93
0.98
0.99
3. A randomly selected student scored above 90.
4. A randomly selected student scored below 75.
5. A randomly selected student scored between 75 and 85
Holt McDougal Algebra 2
Measures of Central Tendency and
Variation
Lesson 2.3 Practice B
Holt McDougal Algebra 2