1. WARM – UP – Calculator
There is a 22% chance of making a “5” on the AP Statistics Exam. In a class of 30, what is the probability that…
…half of the class will get 5’s?
…a third of the class get 5’s?
…at least 6 students get 5’s?
P(x = 15)
= binompdf(30, 0.22, 15) =
P(x = 10)
= binompdf(30, 0.22, 10) =
P(x ≥ 6)
= 1 – P(x ≤ 5)
= 1 – binomcdf(30, 0.22, 5) =

4. Ch. 17 Binomial Distributions
Binomial Probability:
The Probability of having ‘k’ Successes out of ‘n’ Attempts.

5. The Normal Approximation to the Binomial Model
If np ≥ 10 and n(1 – p) ≥ 10 then you can use the Binomial Mean and Standard Deviation with the Normal Model. Mean: µ = np and
Standard Deviation: σ = √np(1 – p)

6. EXAMPLE: Airlines often overbook flights, believing that 5% of passengers fail to show up for flights (95% do show). Suppose a plane will hold 265 passengers, and the airline sells 275 seats. Using the Normal Model find the probability the airline will not have enough seats and passengers get bumped.
P(x ≥ 266) = ?
μ = np
= 275 · 0.95
µ =
σ = √np(1 – p) =
√ 275(0.95)(0.05)
σ =
P(x ≥ 266) =
P(z ≥ 1.314) =
Normalcdf(1.314, E99) =
P(x ≥ 266) =
266
261.25

7. 0.0019 Normal Approximation to Binomial Probability P(x ≤ 50) = µ = np
EXAMPLE: Police estimate that 20% of drivers do not wear their seatbelts. They set up a Safety Roadblock to check for seatbelt usage. If they stop 360 cars, use a Normal Model to find the probability that 50 or less drivers are unbelted.
Normal Approximation to Binomial Probability
P(x ≤ 50) =
µ = np
µ = 360(.20)
µ = 72
σ = √np(1 – p)
σ = √360(.2)(1 – .2)
σ = 7.589
P(x ≤ 50) =
P(z ≤ ) =
Normalcdf(-E99, ) =
0.0019

8. Ch. 17 - The Geometric Distribution (Cont.)
Geometric Probability: The Probability of taking ‘k’ number of Attempts before obtaining the 1st Success.
P(x = k) = (1 – p)(k – 1)·p or (Failure)(k-1)∙(Success)
Geometric Mean
A Geometric Distribution, G(p,k) has mean:
Expected Value = μ = 1/p
EXAMPLE: A baseball player has a probability 0.15 of making a random homerun.
a.) Find the probability that he makes his first homerun on the 4th at bat.
b.) Find the average number of at bats he will need to make his first homerun.
(.85)3∙(.15) =
0.0921
1/0.15 =
6.667

9. HW: Page 399: 13-16, 19, 20

10. e) Use Binom(5, 13)
f) Use Binom(5, 13)

11. HW: Page 399:`` 13-16, 19, 20

15. Geometric Mean µ = 1/p µ = 1/.20 µ = 5 Geometric Prob. P(x=6) (.85x.2)
EXAMPLE: Police estimate that 20% of drivers do not wear their seatbelts. They set up a Safety Roadblock to check for seatbelt usage.
How many cars do they expect to stop before finding a driver who is not using his/her seatbelt?
Geometric Mean
µ = 1/p
µ = 1/.20
µ = 5
2. What is the probability that the first unbelted driver is the 6th car?
Geometric Prob.
P(x=6)
(.85x.2) =
3. What is the probability that the first 10 drivers stopped are wearing their seatbelts?
0.80 wear seat belts P(10 out of 10)
Binomial Prob.
P(x=10)
Binompdf(10,.80,10) OR =
4. If the police stop 30 cars, how many drivers are expected to be unbelted? What is the standard deviation?
µ = 6
Binomial Mean
µ = np
µ = 30(.20)
σ = √np(1 – p)
σ = √30(.2)(1 – .2)
σ =

16. Binomial Prob. P(x= 4) Binompdf(10,.20,4) = .088 Binomial Prob.
EXAMPLE: Police estimate that 20% of drivers do not wear their seatbelts. They set up a Safety Roadblock to check for seatbelt usage.
5. What is the probability that the police will find 4 unbelted drivers within the first 10 drivers stopped?
Binomial Prob.
P(x= 4)
Binompdf(10,.20,4)
= .088
6. What is the probability that the police will find at least 4 unbelted drivers within the first 10 drivers stopped?
Binomial Prob.
P(x≥ 4)
= 1 – P(x ≤ 3)
1 - Binomcdf(10,.20,3) =

17. WARM – UP
A college basketball player has an 82% probability of making Free throws. During a typical game he usually attempts 12 free throws. Find:
The prob. of making exactly 5 baskets.
2. The prob. of making at most 5 baskets.
3. The prob. of making at least 9 baskets.
P(x = 5)
P(x = 5)
= binompdf(12, 0.82, 5) =
P(x ≤ 5)
= binomcdf(12, 0.82, 5) =
P(x ≥ 9)
= 1 – P(x ≤ 8)
= 1 – binomcdf(12, 0.82, 8) =

18. WARM – UP
2003 #3

19. Solution
Part (a):