Unit 5 Chapter 6 Summary



Unit 4 Chapter 4 Summary

4.1 Introduction to Random Variables and Probability Distribution

1. Discrete Random Variable occurs when the observations of a quantitative random variable can take on only a finite number of values or a countable number of values.

2. Continuous Random Variables occurs when the observations of a quantitative random variable can take on any countless numbers of values in a line interval.

Examples of Discrete Random Variables:

1. The cost of conducting a medical experiment

2. The number of supermodels who ate chocolate cake yesterday

3. The number of Business professors who read the Wall Street Journal each day

4. The number of Bald Eagles located in New York State

Examples of Continuous Random Variables:

1. The exact life span (age) of a puppy

2. The weight of a feather

3. The height of a randomly selected elephant in Africa

4. The exact time it takes to calculate 786 – 67

A Probability Distribution is the assignment of probabilities to all specific values of a random variable. A probability is assigned to each value. The probabilities should be between 0 and 1. The sum of all the probabilities must equal 1.

Identify which is a probability distribution.

|a. |b. |

|x |x |

|P(x) |P(x) |

| | |

|0 |0 |

|0.4219 |0.502 |

| | |

|1 |1 |

|0.4219 |0.365 |

| | |

|2 |2 |

|0.1406 |0.098 |

| | |

|3 |3 |

|0.0156 |0.011 |

| | |

|Sum |4 |

|1 |0.001 |

| | |

| |Sum |

|This is probability distribution since the values are between 0 |0.977 |

|and 1 and the sum is 1. | |

| | |

| |This has probabilities between 0 and 1 but the sum is not equal |

| |to 1. Therefore b is not a probability distribution. |

The mean and standard deviation of a discrete probability distribution are:

Mean [pic]

Standard deviation [pic]

For the probability distribution a, Excel is used to set up columns and the totals as shown.

| | | | | | |

|x |P(x) |x•P(x) |x - µ |(x - µ)^2 |(x - µ)^2 •P(x) |

|0 |0.4219 |0 |-0.7499 |0.5624 |0.2373 |

|1 |0.4219 |0.4219 |0.2501 |0.0626 |0.0264 |

|2 |0.1406 |0.2812 |1.2501 |1.5628 |0.2197 |

|3 |0.0156 |0.0468 |2.2501 |5.0630 |0.0790 |

|Sum |1 |0.7499 | | |0.5623 |

| | | | | | |

|Mean µ = ∑ (x•P(x)) = 0.7499 | | | |

| | | | | | |

|Standard Deviation σ = SQRT [ (x - µ)^2 • P(x)] = |0.7499 |

Note: It is a coincidence that µ and σ are same in this example.

4.2 Binomial Distribution

Features of a binomial experiment

1. There are a fixed number of trials denoted by n.

2. The n trials are independent and performed under identical conditions.

3. Each trial has only two outcomes: success denoted by S and failure denoted by F

4. For each trial the probability of success is the same and denoted by p. The probability of failure is denote by q and p + q=1 (or q = 1 - p)

5. The central problem is to determine the probability of r successes out of n trials. P(r) = ?

Probability for the binomial distribution (probability of r successes out of n trials)

[pic]

Where r = number of successes

n = number of trials

p = probability of one success

q = probability of one failure

The Table 2 is a simpler way to compute the binomial probability. If you have n, r, and p you can use this table to get P(r). Locate the n group, then the p column. Read the table for r.

For n and r, find the specific probability:

1. n=3, r=0, p=0.05 P(r=0) = 0.857

2. n=15, r=12, p=0.9 P(r=12) = 0.129

3. n=10, r ≤ 4, p=0.6

P(r ≤ 4)= P(0) + P(1) + P(2) + P(3) + P(4)

= .000+.002+.011+.042+.111 = 0.166

4. n=10, r ≥ 8, p = 0.75

P(r ≥ 8) = P(8) + P(9) + P(10)

= .282 + .188 + .056 = 0.526

5. n=5, r is at least 4 (same as 4 and 5), P =0.8

P(at least 4) = P(4) + P(5) = .410 + .328 = 0.738

6. n = 8, r is less than 3 (same as 0,1, 2), p =0.8

P(r < 3) = P(0) + P(1) + P(2) = .000 + .000 + .001 += 0.001

7. The TV called Woops has been successful for many years. The show last season had a share of 20, meaning among the TV sets in use, 20% were tuned to Woops. Assume that an advertiser wants to verify that 20% share value by conducting its own survey. A pilot survey of 10 households having TV sets in use at the same time as the broadcast of Woops. (n = 10, p = 0.2)

a. Find the probability that none of the households are tuned to Woops. P(r= 0) = 0.107

b. Find the probability that at least one of the households are tuned to Woops. P(at least 1) = 1 – P(0) = 1 - .107 = .893

c. Find the probability that at most one of the households are tuned to Woops. P(at most 1) = P(0) + P(1) = .107 + .268 = .375

8. A blind marksman finds that on the average he hits the target 4 out of 5 times. If he fires 4 shots, find the probability of

n = 5, p = 4/5 = 0.8 (hits), q = 1 – 0.8 = 0.2 (misses)

d. more than 2 hits

Since problem asks for hits, use 0.8 for the probability.

P(r > 2) = P( r ( 3 ) = P(3) + P(4) + P(5)

= 0.205 + 0.410 + 0.328 = .943

e. At least 3 misses

Since problem asks for misses, use 0.2 for the probability.

P( r ( 3 ) = P(3) + P(4) + P(5)

= 0.051 + 0.006 + 0.000 = 0 .057

Graphs can be produced for each group and probability on Table 2.

| | | |

|Table 2 for n = 5, p = 0.6 |

|r |P (r = ) | |

|0 |0.010 | |

|1 |0.077 | |

|2 |0.230 | |

|3 |0.346 | |

|4 |0.259 | |

|5 |0.078 | |

| | | |

[pic]

For a binomial distribution, compute µ and σ .

Mean µ = np n is number of trials

Standard deviation [pic] p probability of success

q = 1 – p probability of failure

For n = 5 and p = 0.6

Mean µ = np = 5 ( 0.6) = 3

Standard deviation [pic]

Other examples:

n = 12, p = 0.8 q = 1 – p = 1 – 0.8 = 0.2

µ = 12(0.8) = 9.6

[pic]

Mars, Inc. claims that 20% of its M&M plain candies are orange. A random sample of 100 is selected. Find the mean and standard deviation of the number of orange candies in one such 100 group.

n = 100 p = 0.2 q = 1 – 0.2 = 0.8

µ = 100 (0.2) = 20

[pic]

1. n=5, r=0, p=0.05

| |[pic] |

|P(0) = [pic] | |

2. n=5, r=2, p=0.05

|P(2) = |[pic] |

|[pic] | |

| | |

|(The slight difference in answer is due to rounding) | |

3. n= 5, p = .05, r ( 5

|n= 5, p = .05 |[pic] |

|Find P(r (5) = | |

|P(0) + P(1) + P(2) + P(3) + P(4) + P(5) | |

|= .774 + .204 + .021 + .001 + .000 + .000 | |

|= 1 | |

4. n = 3, p = .45 r > 1

|P(r > 1) = P( r ( 2 ) = P(2) + P(3) = .334 + .091 (.425 |

|Or 1 – P( r ( 1) = 1 – (.166 + .408) ( .426 (difference due to rounding in table) |

|[pic] |

5. n = 3, p = .45 , at least 1 for r

|P( r ( 1 ) = P(1) + P(2) + P(3) = .408 + .334 + .091 = .833 |

| |

|[pic] |

Examples using the full table

|6. n=5, r is at least 4 (same as 4 and 5), P =0.8 |

|P(at least 4) = P(4) + P(5) = .410 + .328 = 0.738 |

| |

|7. n=15, r=12, p=0.9 P(r=12) = 0.129 |

| |

|8. n=10, r ≤ 4, p=0.6 |

|P(r ≤ 4)= P(0) + P(1) + P(2) + P(3) + P(4) |

|= .000+.002+.011+.042+.111 = 0.166 |

| |

|9. n=10, r ≥ 8, p = 0.75 |

|P(r ≥ 8) = P(8) + P(9) + P(10) |

|= .282 + .188 + .056 = 0.526 |

| |

|10. n=5, r is at least 4 (same as 4 and 5), P =0.8 |

|P(at least 4) = P(4) + P(5) = .410 + .328 = 0.738 |

| |

|11. n = 8, r is less than 3 (same as 0,1, 2), p =0.8 |

|P(r < 3) = P(0) + P(1) + P(2) = .000 + .000 + .001 += 0.001 |

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