Chapter Five. The Binomial Probability Distribution and Related Topics

Chapter Five The Binomial Probability Distribution and Related Topics

Section 3 Additional Properties of the Binomial Distribution

Essential Questions How are the mean and standard deviation determined when using a binomial distribution? What must be true for this process to be applied to the distribution?

Student Objectives The student will determine the values of a binomial distribution. The student will construct a histogram for a binomial distribution. The student will compute the minimum number of trials needed to achieve a given probability of success. The student will determine the least number of trials necessary to have a required value of µ. The student will determine the minimum number of trials (quota) required to reach a given percentage of success. The student will determine a range of unusual values of the a binomial experiment.

Key Terms Histogram Mean Quota Standard Deviation Unusual values Variance

Mean and Standard Deviation of a Binomial Distribution Mean = µ µ = np Standard Deviation = σ σ = npq

Mean and standard deviation of the binomial distribution Find the mean and standard deviation of the probability distribution for tossing four coins and observing the number of heads appearing. n = 4 p = 0.5 q = 1 p q = 1 0.5 q = 0.5

Mean and standard deviation of the binomial distribution Mean Standard Deviation µ = np σ = npq µ = 4(.5) σ = 4(.5)(.5) µ = 2 σ = 1 σ =1

Find the mean and standard deviation: A marksman takes eight shots at a target. He normally hits the target 70% of the time. Find the mean number of times he hits the target and standard deviation of hits. n = 8 p = 0.7 q = 1-0.7 q = 0.3 µ = np σ = npq µ = 8(.7) σ = 8(.7)(.3) µ = 5.6 σ = 1.68 σ = 1.2961

Sample Problems: 1. Eight trees are going to be planted in the northern corner of the new track area. Determine the mean number of trees that expected to survive if the probability of survival for each tree is 0.85? What is the standard deviation? µ = np µ = ( 8) ( 0.85) µ = 6.8 σ = npq σ = ( 8) ( 0.85) ( 0.15) σ = 1.02 σ = 1.0100

Sample Problems: 2. The probability of having a male child is 0.55. If a family is planning on having seven children. What is the mean number of boys the family can expect? What is the standard deviation? µ = np σ = npq µ = ( 7) ( 0.55) µ = 3.85 σ = ( 7) ( 0.55) ( 0.45) σ = 1.7325 σ = 1.3162

?Application Questions? How do you find the minimum number of trials to reach a given value for µ?

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to hit the target an average of 9 times? µ = 9 and p = 0.7 µ = np 9 = n( 0.7) 12.8571 = n 13 = n So, the marksman would need to take 13 shots to have 9 as the mean number of times that he hits the target. Remember to ALWAYS round your answer upward unless it is an integer value.

Sample Problems: 3. What is the least number of children the parents should plan to have if they wish to be 95% certain that they have at least 4 boys? 0.95 1.05 1 binomcdf (n, 0.55, 3) ( ) ( ) 1 P r 3 P r 4 n = 12 P( r 4) 1 P( r 3) 1 binomcdf (12, 0.55, 3) 1 0.03557 0.9644 The parents should plan to have 12 children to be 95% certain that they have at least 4 boys.

?Application Questions? How do you find the minimum number of trials to be P% sure that you reach a given value for µ?

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to be 95% certain that he hits the target at least 6 times? If n = 7 and p = 0.70 P( r 6) 1 P( r 5) 1 binomcdf ( 7, 0.7, 5) 1 0.6705 0.3294 Too low!

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to be 95% certain that he hits the target at least 6 times? If n = 8 and p = 0.70 P( r 6) 1 P( r 5) 1 binomcdf ( 8, 0.7, 5) 1 0.4482 0.5518 Too low!

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to be 95% certain that he hits the target at least 6 times? If n = 9 and p = 0.70 P( r 6) 1 P( r 5) 1 binomcdf ( 9, 0.7, 5) 1 0.2703 0.7297 Too low!

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to be 95% certain that he hits the target at least 6 times? If n = 10 and p = 0.70 P( r 6) 1 P( r 5) 1 binomcdf ( 10, 0.7, 5) 1 0.1503 0.8497 Too low!

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to be 95% certain that he hits the target at least 6 times? If n = 11 and p = 0.70 P( r 6) 1 P( r 5) 1 binomcdf ( 11, 0.7, 5) 1 0.07822 0.9217 Too low!

A marksman takes eight shots at a target. He normally hits the target 70% of the time. How many shots must he take if he wishes to be 95% certain that he hits the target at least 6 times? If n = 12 and p = 0.70 P( r 6) 1 P( r 5) 1 binomcdf ( 12, 0.7, 5) 1 0.03860 0.9614 Finally!

Sample Problems: 4. What is the least number of children the parents should plan to have if they wish to be 99.99% certain that they have at least 4 boys? P r 4 ( ) 1 P( r 3) 1 binomcdf (22, 0.55, 3) 1 0.0007477 0.9999 The parents should plan to have 22 children to be 99.99% certain that they have at least 4 boys.

?Application Questions? How do you find the range for unusual data values for a binomial distribution?

Unusual Values For a binomial distribution, it is unusual for the number of successes r to be lower or higher than the inequalities stated below. r < µ 2.5σ or r > µ + 2.5σ Remember that r is the number of success and r must be an integer value in the range given below. 0 r n

Find the values of r that would be considered unusual: A marksman takes eight shots at a target. He normally hits the target 70% of the time. Find the mean number of times he hits the target and standard deviation of hits. n = 8 p = 0.7 q = 1-0.7 q = 0.3 µ = np σ = npq µ = 8(.7) σ = 8(.7)(.3) µ = 5.6 σ = 1.68 σ = 1.2961

Find the values of r that would be considered unusual: A marksman takes eight shots at a target. He normally hits the target 70% of the time. Find the mean number of times he hits the target and standard deviation of hits. µ = 5.6 σ = 1.2961 r < 5.6 2.5( 1.2961) OR r > 5.6 + 2.5( 1.2961) r < 5.6 3.2404 r > 5.6 + 3.2404 r < 2.3596 r > 8.8404 The number of successes that would be unusual for the marksman would be to hit the target 0, 1, or 2 times.

Sample Problems: 5. The probability that it will snow any day during the Farm Show week (7 days) is 0.37. What is the mean number of days it can be expected to snow? What is the standard deviation? Determine the number of days that would be considered unusual if it snowed that number of days? µ = 2.59 σ = 1.2773 r < 2.59 2.5( 1.2774) OR r > 2.59 + 2.5( 1.2774) r < 2.59 3.1935 r > 2.59 + 3.1935 r < 0.6035 r > 5.7835 The number of successes that would be unusual for the number of days that it would snow during Farm Show week is 6 and 7.

?Application Questions? How do you construct a histogram from a binomial probability distribution?

Constructing a Histogram for a Binomial Probability Distribution A marksman takes eight shots at a target. He normally hits the target 70% of the time. Find the mean number of times he hits the target and standard deviation of hits. µ = np Step 1. Complete a probability chart. r ( ) P r ( ) r P r 0 0.0001 0.0000 1 0.0012 0.0012 2 0.0100 0.0200 3 0.0467 0.1400 4 0.1361 0.5445 5 0.2541 1.2706 6 0.2965 1.7789 7 0.1977 1.3836 8 0.0576 0.4612 Total 1.0000 5.6000

Constructing a Histogram for a Binomial Probability Distribution A marksman takes eight shots at a target. He normally hits the target 70% of the time. Find the mean number of times he hits the target and standard deviation of hits. Step 2. Place the r values on the horizontal axis. Step 3. Place the P(r) values on the vertical axis. Step 4. Construct a over each r value extending from r - 0.5 to r + 0.5. The height of the bar corresponds to P(r). Probability 0.30 0.25 0.20 0.15 0.10 0.05 0.00 A Marksman s Success 0.2965 0.2541 0.1977 0.1361 0.0576 0.0467 0.0100 0.0001 0.0012 0 1 2 3 4 5 6 7 8 Number of Hits

Sample Problems: 6. A chart showing the number of times that a student will purchase the main meal from the cafeteria is given below. If the probability of purchasing the main meal is 70%, complete the chart and then complete the histogram on the given grid. Determine the mean number of meals purchased by a student. What is the standard deviation? How many people would you need to survey to be 97% certain that you have at least 5 people that purchase the main meal on any given day? µ = np σ = npq µ = 5(.7) σ = 5(.7)(.3) µ = 3.5 σ = 1.05 σ = 1.0247

Sample Problem Constructing a Histogram for a Binomial Probability Distribution A chart showing the number of times that a student will purchase the main meal from the cafeteria is given below. If the probability of purchasing the main meal is 70%, complete the chart and then complete the histogram on the given grid. Determine the mean number of meals purchased by a student. What is the standard deviation? How many people would you need to survey to be 97% certain that you have at least 5 people that purchase the main meal on any given day? r P( r) r P( r) 0 0.0024 0.0000 1 0.0283 0.0283 2 0.1323 0.2646 3 0.3087 0.9261 µ = np Step 1. Complete a probability chart. 4 0.3601 1.4406 5 0.1681 0.8404 Total 1.0000 3.5000

Sample Problem Constructing a Histogram for a Binomial Probability Distribution If the probability of purchasing the main meal is 70%, complete the chart and then complete the histogram on the given grid. Step 2. Place the r values on the horizontal axis. Step 3. Place the P(r) values on the vertical axis. Step 4. Construct a over each r value extending from r - 0.5 to r + 0.5. The height of the bar corresponds to P(r). Probability 0.40 0.30 0.20 0.10 0.00 Buying the Main Meal 0.3602 0.3087 0.1681 0.1323 0.0283 0.0024 0 1 2 3 4 5 Number of Purchases

Sample Problems: 6. How many people would you need to survey to be 97% certain that you have at least 5 people that purchase the main meal on any given day? If n = 6 and p = 0.70 P( r 5) 1 P( r 4) 1 binomcdf ( 11, 0.70, 4) 1 0.02162 0.9784 You would need to survey 11 students to be 97% certain that they have at least 5 students that have purchased the main meal on any given day.