Chapter (7) Continuous Probability Distributions Examples - KSU

[Pages:18]Chapter (7) Continuous Probability Distributions

Examples

The uniform distribution

Example (1) Australian sheepdogs have a relatively short life .The length of their life follows a uniform distribution between 8 and 14 years.

1. Draw this uniform distribution. What are the height and base values? 2. Show the total area under the curve is 1. 3. Calculate the mean and the standard deviation of this distribution. 4. What is the probability a particular dog lives between 10 and 12 years? 5. What is the probability a particular dog lives greater than 10? 6. What is the probability a dog will live less than 9 years? Solution: 1.

The height 1 1 1 0.167 b a 14 8 6

Maximum=14 Minimum=8

2.

The total area 1 14 8 1 6 1

14 8

6

3.

a b 8 14 22 11

2

22

1

 b a2 14 82 62 36 3 1.73

12

12

12 12

4.

p10 X 12 1 12 10 1 2 1 0.3333

14 8

63

5.

pX 10 p10 X 14 1 14 10 1 4 4 0.667

14 8

66

6.

pX 9 p8 X 9 1 9 8 1 1 0.167

14 8

6

2

Normal probability distribution

How to find the area under the normal curve? If 50 & 6 Find

1PX 60.8 P Z 60.8 50

6

P Z 10.8 pZ 1.8 0.5 P0 Z 1.8

6

0.5 0.4641 0.9641

2PX 39.2 P Z 39.2 50

6

P Z 10.8 pZ 1.8 0.5 P1.8 Z 0

6

0.5 0.4641 0.9641

3PX 60.8 P Z 60.8 50

6

P Z 10.8 pZ 1.8 0.5 P0 Z 1.8

6

0.5 0.4641 0.0359

3

4PX 39.2 P Z 39.2 50

6

P Z 10.8 pZ 1.8 0.5 P1.8 Z 0

6

0.5 0.4641 0.0359

5P59 X 60.8 P 59 50 Z 60.8 50

6

6

P 9 Z 10.8 p1.5 Z 1.8 P0 Z 1.8 P0 Z 1.5

6

6

0.4641 0.4332 0.0309

6P39.2 X 41 P 39.2 50 Z 41 50

6

6

P 10.8 Z 9 p1.8 Z 1.5

6

6

P0 Z 1.8 P0 Z 1.5

0.4641 0.4332 0.0309

4

7P41 X 60.8 P 41 50 Z 60.8 50

6

6

P 9 Z 10.8 p1.5 Z 1.8 P0 Z 1.8 P1.5 Z 0

6

6

0.4641 0.4332 0.8973

5

Example (2)

Let X be a normally distributed random variable with mean 65 and standard

deviation 13. Find the standard normal random variable (z) for P( X>80)

Solution:

P(X 80) P(Z 80 65) P(Z 15) P(Z 1.15)

13

13

0.5 0.3749 0.1251

Example (3)

If the mean = 65 and standard deviation =13. Find x from the following:

1. z = 0.6

2. z = -1.93

Solution:

1. z 0.6

x 65 0.613 65 7.8 72.8

2. z 1.93 x 65 1.9313 65 25.09 39.91

The Empirical Rule

Example (4) A sample of the rental rates at University Park Apartments approximates a systematical, bell- shaped distribution. The sample mean is $500; the standard deviation is $20.Using the Empirical Rule, answer these questions: About 68 percent of the monthly food expenditures are between what two amounts? 1. About 95 percent of the monthly food expenditures are between what two amounts? 1. About all of the monthly (99.7%) food expenditures are between what two amounts?

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Solution

1- X 1 500 120 500 20

$480,$520

About 68 percent are between $480 and $520.

2- X 2 500 220 500 40

$460,$540

About 95 percent are between $460 and $540.

3- X 2 500 320 500 60

$440,$560

About 99.7 percent are between $440 and $560.

Example (5) The mean of a normal probability distribution is 120; the standard deviation is 10.

a. About 68 percent of the observations lie between what two values? b. About 95 percent of the observations lie between what two values? c. About 99 percent of the observations lie between what two values? Solution:

a. 1 120 110 120 10

130 and 110

b. 2 120 210 120 20

140 and 100

c. 3 120 310 120 30

150 and 90

Example (6) Studies show that gasoline use for compact cars sold in the United States is normally distributed, with a mean of 25.5 miles per gallon (mpg) and a standard deviation of 4.5 mpg. Find the probability of compact cars that get: 1. 30 mpg or more. 2. 30 mpg or less. 3. Between 30 and 35. 4. Between 30 and 21.

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Solution:

25.5 4.5

1. Px 30 P(Z 30 25.5) P(Z 4.5) Pz 1 0.5 1

4.5

4.5

0.5 0.3413 0.1587

2. Px 30 P Z 30 25.5 P Z 4.5 Pz 1 0.5 1

4.5 4.5

0.5 0.3413 0.8413

3. P30 x 35 P 30 25.5 Z 35 25.5 P 4.5 Z 9.5

4.5

4.5 4.5

4.5

P1 z 2.11 2.11 1 0.4826 0.3413 0.1413

4. P21 x 30 P 21 25.5 Z 30 25.5 P 4.5 Z 4.5

4.5

4.5 4.5

4.5

P1 z 1 1 1 0.3413 0.3413 0.6826

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