ppt04. Computing the Mean of a Discrete Probability Distribution.pptx

Statistics and Probability

1. Computing the Mean of a Discrete Probability Distribution

2. Lesson Objectives At the end of this lesson, you should be able to: illustrate and calculate the mean of a discrete random variable; interpret the mean of a discrete random variable; and solve problems involving mean of probability distributions.

3. Content Entry Card: Summation Sample Problem 1: Number of Spots Formula for the Mean of the Probability Distribution Sample Problem 2: Grocery Items Sample Problem 3: Surgery Patients Seatwork Enrichment Assessment

4. Entry Card Given the values of the variables X and Y, evaluate the following summations. 1. 2. 3. 4. 5.  

5. Problem 1: Number of Spots Consider rolling a die. What is the average number of spots that would appear? Solution: Step 1: Construct the probability distribution for the random variable X representing the number of spots that would appear. Step 2: Multiply the value of the random variable X by the corresponding probability. Step 3: Add the results obtain in Step 2.

6. Step 1 Number of Spots X Probability P(X) 1 2 3 4 5 6 Number of Spots X Probability P(X) 1 2 3 4 5 6

7. Number of Spots X Probability P(X) X ∙ P(X) 1 2 3 4 5 6 Number of Spots X Probability P(X) X ∙ P(X) 1 2 3 4 5 6 Step 2

8. Step 3 Number of Spots X Probability P(X) X ∙ P(X) 1 2 3 4 5 6 Number of Spots X Probability P(X) X ∙ P(X) 1 2 3 4 5 6

9. Formula for the Mean of the Probability Distribution The mean of a random variable with a discrete probability distribution is: or where: are the values of the random variable X; and , , , …, are the corresponding probabilities.  

10. Problem 2: Grocery Items The probabilities that a costumer will buy 1, 2, 3, 4, or 5 items in a grocery store are , respectively. What is the average number of items that a costumer will buy?  

11. Step 1 Number of Items X Probability P(X) 1 2 3 4 5 Number of Items X Probability P(X) 1 2 3 4 5

12. Number of Items X Probability P(X) X ∙ P(X) 1 2 3 4 5 Number of Items X Probability P(X) X ∙ P(X) 1 2 3 4 5 Step 2

13. Step 3 Number of Items X Probability P(X) X ∙ P(X) 1 2 3 4 5 Number of Items X Probability P(X) X ∙ P(X) 1 2 3 4 5

14. Problem 3: Surgery Patients The probabilities that a surgeon operates on 3, 4, 5, 6, or 7 patients in any day are 0.15, 0.10, 0.20, 0.25, and 0.30, respectively. Find the average number of patients that a surgeon operates on a day.

15. Step 1 Number of Patients X Probability P(X) 3 0.15 4 0.10 5 0.20 6 0.25 7 0.30

16. Number of Patients X Probability P(X) X ∙ P(X) 3 0.15 0.45 4 0.10 0.40 5 0.20 1.00 6 0.25 1.50 7 0.30 2.10 Step 2

17. Step 3 Number of Patients X Probability P(X) X ∙ P(X) 3 0.15 0.45 4 0.10 0.40 5 0.20 1.00 6 0.25 1.50 7 0.30 2.10 Number of Patients X Probability P(X) X ∙ P(X) 3 0.15 0.45 4 0.10 0.40 5 0.20 1.00 6 0.25 1.50 7 0.30 2.10

18. Complete the table below and find the mean of the following probability distribution . E( x.Px )= 6.75 X P(X) X ∙ P(X) 3 0.15 0.45 6 0.35 2.10 8 0.40 3.20 10 0.10 1.00

19. X P(X) X ∙ P(X) 1 1/7 6 6/7 11 33/7 16 16/7 21 21/7 =77/7 or 11 X P(X) X ∙ P(X) 1 1/7 6 6/7 11 33/7 16 16/7 21 21/7 =77/7 or 11 2.

20. 3. Number of Items X Probability P(X) X ∙ P(X) 1 4/9 3 2/3 or 6/9 5 5/9 7 14/9= 29/9 or 3.22 Number of Items X Probability P(X) X ∙ P(X) 1 4/9 3 2/3 or 6/9 5 5/9 7 14/9= 29/9 or 3.22

21. Find the mean of the probability distribution of the random variable X, which can take only values 1, 2, and 3, given that .   1 10/33 10/33 2 1/3 2/3 3 12/33 36/33 =68/33 or 2.06

22. Questions? Clarifications?

23. Thank You for listening . . .