1. 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: 1.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. 4.The prob. he will make his first basket on the 3rd throw. P(x = 5) = binompdf(12, 0.82, 5) = 0.0018 P(x ≤ 5) = binomcdf(12, 0.82, 5) = 0.0021 P(x = 3) = geometpdf(0.82, 3) = 0.0265 P(x ≥ 9) = 1 – binomcdf(12, 0.82, 8) = 0.8448 = 1 – P(x ≤ 8) P(x = 5)

2. Example #2 There is a 22% chance of making a “5” on the AP Statistics Exam. In a class of 30, what is the probability that… (a)…half of the class will get 5’s? (b)…a third of the class get 5’s? (c)…at least 6 students get 5’s? P(x = 15) = binompdf(30, 0.22, 15) = 0.000511 P(x = 10) = binompdf(30, 0.22, 10) = 0.0554 P(x ≥ 6) = 1 – binomcdf(30, 0.22, 5) = 0.6739 = 1 – P(x ≤ 5)

3. Binomial Mean and Standard Deviation A Binomial Distribution, B (n, p) has: Mean = μ = np Standard Deviation = σ = √ np(1 – p) EXAMPLE: A college basketball player has an 82% probability of making Free throws. During a typical game he usually attempts 12 free throws. Find: 1.The mean number of baskets he is expected to make. 2.The Standard Deviation of the number of baskets. μ = np = 12 · 0.82 σ = √ np(1 – p) = √ 12(0.82)(1 – 0.82) μ = 9.84 σ = 1.331

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

5. 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 µ = np µ = 360(.20) σ = 7.589 σ = √ np(1 – p) σ = √ 360(.2)(1 –.2) µ = 72 P(x ≤ 50) = Normalcdf(-E99, -2.899) = P(z ≤ -2.899) = 0.0019 P(x ≤ 50) =

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. μ = np √ 275(0.95)(0.05) σ = 3.6142 P(x ≥ 266) = Normalcdf(1.314, E99) = P(x ≥ 266) = 0.0944 µ = 261.25 P(x ≥ 266) = ? σ = √ np(1 – p) = P(z ≥ 1.314) = = 275 · 0.95 261.25 266

7. EXAMPLE: Police estimate that 20% of drivers do not wear their seatbelts. They set up a Safety Roadblock to check for seatbelt usage. 1.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 6 th car? Geometric Prob. P(x=6) Geometpdf(.2,6) = 0.0655 3.What is the probability that the first 10 drivers stopped are wearing their seatbelts? Binomial Prob. P(x=10) Binompdf(10,.80,10) = 0.1074 0.80 wear seat belts P(10 out of 10) OR.80 10 4.If the police stop 30 cars, how many drivers are expected to be unbelted? What is the standard deviation? Binomial Mean µ = np µ = 30(.20) σ = 2.1909 σ = √ np(1 – p) σ = √ 30(.2)(1 –.2) µ = 6

8. 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? 6. What is the probability that the police will find the first unbelted driver within the first 10 drivers stopped Geometric Prob. P(x≤10) Geometcdf(.2,10) = 0.8926 Binomial Prob. P(x= 4) Binompdf(10,.20,4) =.088 7.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 - Binomcdf(10,.20,3) = 0.1209 = 1 – P(x ≤ 3)

9. EXAMPLE: Shortly after the introduction of the Euro coin, newspapers around the world published articles claiming that the coin was biased. If someone spun the coin 250 times, the coin landed heads up 140 times (56%). Do YOU think this is evidence that the coin is weighted? Normal Approximation to Binomial Probability µ = np µ = 250(.50) σ = 7.906 σ = √ np(1 – p) σ = √ 250(.5)(1 –.5) µ = 125 P(x ≥ 140) = Normalcdf(1.897, E99) = P(z ≥ 1.897) = 0.0289 P(x ≥ 140) =

11. WARM – UP 2003 #3

12. Solution Part (a):