1. Section 6.2 Binomial Distribution
Day 2

2. How do you decide whether or not you are dealing with a binomial distribution?

3. How do you decide whether or not you are dealing with a binomial distribution?
Check BINS

4. B: a series of trials must be binomial
BINS
B: a series of trials must be binomial
-- each trial must have one of two different outcomes, one called a “success” and the other a “failure”

5. BINS B: a series of trials must be binomial
I: each trial is independent of the others
-- one event does not affect the other
-- sampling with replacement
-- small sample from a large population

6. BINS
B: a series of trials must be binomial
I: each trial is independent of the others
N: there is a fixed number, n, of trials

7. BINS
B: a series of trials must be binomial
I: each trial is independent of the others
N: there is a fixed number, n, of trials
S: the probability, p, of a success is the same on each trial, with 0 < p <1

8. The proportion of households in the United States that own three or more vehicles is approximately 0.16.
Suppose you pick eight households at random and count the number of households that own three or more vehicles.

9. The proportion of households in the United States that own three or more vehicles is approximately 0.16.
Suppose you pick eight households at random and count the number of households that own three or more vehicles.
Verify that you can treat your random sample as a binomial situation.

10. B: two possible outcomes
-- household either owns three or more vehicles or does not

11. The proportion of households in the United States that own three or more vehicles is approximately 0.16.
Suppose you pick eight households at random and count the number of households that own three or more vehicles.

12. I: sample size (8) is small compared to the population of households in the United States so the households are independent of each other

13. The proportion of households in the United States that own three or more vehicles is approximately 0.16.
Suppose you pick eight households at random and count the number of households that own three or more vehicles.

14. N: fixed number of trials, n = 8

15. The proportion of households in the United States that own three or more vehicles is approximately 0.16.
Suppose you pick eight households at random and count the number of households that own three or more vehicles.

16. S: P(success) = 0.16 for each trial

17. The proportion of households in the United States that own three or more vehicles is approximately Suppose you pick eight households at random and count the number of households that own three or more vehicles.
What is the probability that none of the households in your sample own three or more vehicles?

18. What is the probability that none of the households in your sample own three or more vehicles?
P (X = 0) = binompdf(8, .16, 0) =

19. What is the probability that at least one of the households in your sample own three or more vehicles?

20. What is the probability that at least one of the households in your sample own three or more vehicles?
P (X 1) =

21. What is the probability that at least one of the households in your sample own three or more vehicles?
P (X 1) = 1 – P(X = 0)

22. What is the probability that at least one of the households in your sample own three or more vehicles?
P (X 1) = 1 – P(X = 0) = =

23. Make a probability distribution table for the number of households in your sample that own three or more vehicles.

24. Make a probability distribution table for the number of households in your sample that own three or more vehicles.
binompdf(8, .16)

25. 3 or more vehicles Probability, p 0 0.2479 1 0.3777 2 0.2518 3 0.0959
# households that own
3 or more vehicles Probability, p
Do not use 3.3E-4

26. What is the probability that 5 or fewer households in your sample own three or more vehicles?

27. 3 or more vehicles Probability, p 0 0.2479 1 0.3777 2 0.2518 3 0.0959
# households that own
3 or more vehicles Probability, p
Do not use 3.3E-4

28. What is the probability that 5 or fewer households in your sample own three or more vehicles?
P(X = 5 or fewer) = =

29. P(X = 5 or fewer) = binomcdf(8, .16, 5) 0.9997

30. # households that own binompdf binomcdf
3 or more vehicles Probability, p

31. When can we use the normal probability distribution to represent a binomial situation?

32. When can we use the normal probability distribution to represent a binomial situation?
Recall, as n increases, the shape of the probability distribution becomes more normal.

33. Rule of Thumb
We can use the normal approximation when n and p satisfy both:
(1) np
(2) n(1 – p) 10

34. Shape, Center, and Spread
Center is the expected value (the mean).
E(X) = x = np
Spread is the standard deviation.
x =

35. Page 91

36. Area Under Normal Curve
normalcdf(left bound, right bound, mean, standard deviation)

37. A survey of drivers in the United States found that 15% never use a cell phone while driving. Suppose that drivers arrive at random at an auto inspection station.
(a) If the inspector checks 10 drivers, what is the probability that at least one driver never uses a cell phone while driving?

38. A survey of drivers in the United States found that 15% never use a cell phone while driving. Suppose that drivers arrive at random at an auto inspection station.
(a) If the inspector checks 10 drivers, what is the probability that at least one driver never uses a cell phone while driving?
P(X ≥ 1) = 1 – P(X = 0) =

39. A survey of drivers in the United States found that 15% never use a cell phone while driving. Suppose that drivers arrive at random at an auto inspection station.
(a) If the inspector checks 10 drivers, what is the probability that at least one driver never uses a cell phone while driving?
P(X ≥ 1) = 1 – P(X = 0) =
1- binompdf(10, .15, 0) =

40. 15% never use a cell phone while driving
(b) Suppose the inspector checks 1000 drivers. Use the normal approximation to the binomial distribution to find the approximate probability that at least 13% of these drivers never use a cell phone while driving.

41. 15% never use a cell phone while driving
(b) Suppose the inspector checks 1000 drivers. Use the normal approximation to the binomial distribution to find the approximate probability that at least 13% of these drivers never use a cell phone while driving.
normalcdf(left bound, right bound, mean, standard deviation)

42. 15% never use a cell phone while driving
(b) Suppose the inspector checks 1000 drivers. Use the normal approximation to the binomial distribution to find the approximate probability that at least 13% of these drivers never use a cell phone while driving.
normalcdf(left bound, right bound, mean, standard deviation)
mean = np = 1000(.15) = 150
standard deviation =

43. 15% never use a cell phone while driving
(b) Suppose the inspector checks 1000 drivers. Use the normal approximation to the binomial distribution to find the approximate probability that at least 13% of these drivers never use a cell phone while driving.
normalcdf(left bound, right bound, mean, standard deviation)
normalcdf(130, 1000, 150, ) =

44. 15% never use a cell phone while driving
(b) Suppose the inspector checks 1000 drivers. Use the normal approximation to the binomial distribution to find the approximate probability that at least 13% of these drivers never use a cell phone while driving.
normalcdf(130, 1000, 150, ) =
0.9617

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48. Questions?