Download presentation
Presentation is loading. Please wait.
Published byCoral Gallagher Modified over 9 years ago
1
Chapter 6 The Normal Distribution Section 6-3 The Standard Normal Distribution
2
0.75 0 Find the area under the normal distribution curve. Section 6-3 Exercise #7
3
0 0.79 1.28 Section 6-3 Exercise #15
4
0 2.83 Section 6-3 Exercise #31
5
Find the z value that corresponds to the given area. 0.8962 – 0.5 = 0.3962 0 z 0.8962 Section 6-3 Exercise #45
6
Find the z value that corresponds to the given area. 0 z 0.8962
7
Section 6-4 Applications of the Normal Distribution Chapter 6 The Normal Distribution
8
a. Greater than 700,000. b. Between 500,000 and 600,000. Section 6-4 Exercise #3
9
a. Greater than 700,000 01.63
10
b. Between 500,000 and 600,000.
11
– 2.36 – 0.36 b. Between 500,000 and 600,000.
12
The average credit card debt for college seniors is $3262. If the debt is normally distributed with a standard deviation of $1100, find these probabilities. a.That the senior owes at least $1000 b.That the senior owes more than $4000 c.That the senior owes between $3000 and $4000 Section 6-4 Exercise #11
13
a. That the senior owes at least $1000 – 2.06
14
b. That the senior owes more than $4000 0.67
15
c. That the senior owes between $3000 and $4000. 0.67 – 0.24
16
Section 6-4 Exercise #27
17
The bottom 18% means that 32% of the area is between z and 0. The corresponding z score will be. $18,840.48$24,596 0.32 0.18
18
Section 6-5 The Central Limit Theorem Chapter 6 The Normal Distribution
19
Section 6-5 Exercise #13
20
$2.00 $2.02
21
The average time it takes a group of adults to complete a certain achievement test is 46.2 minutes. The standard deviation is 8 minutes. Assume the variable is normally distributed. Section 6-5 Exercise #21 Average time = 46.2 minutes, Standard deviation = 8 minutes, variable is normally distributed.
23
4346.2
24
4346.2
25
Yes, since it is within one standard deviation of the mean. It is very unlikely, since the probability would be less than 1%.
26
The average cholesterol of a certain brand of eggs is 215 milligrams, and the standard deviation is 15 milligrams. Assume the variable is normally distributed. Section 6-5 Exercise #23
27
220215
28
220
29
Section 6-6 The Normal Approximation to The Binomial Distribution Chapter 6 The Normal Distribution
30
Two out of five adult smokers acquired the habit by age 14. If 400 smokers are randomly selected, find the probability that 170 or more acquired the habit by age 14. Section 6-6 Exercise #5
31
169.5 160
32
The percentage of Americans 25 years or older who have at least some college education is 50.9%. In a random sample of 300 Americans 25 years and older, what is the probability that more than 175 have at least some college education? Section 6-6 Exercise #7
33
175.5 152.7
34
Women comprise 83.3% of all elementary school teachers. In a random sample of 300 elementary school teachers, what is the probability that more than 50 are men? Section 6-6 Exercise #11
35
50.5 50.1
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.