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)P(x = 5) = binompdf(12, 0.82, 5) = = binompdf(12, 0.82, 5) = 0.00180.0018

P(x ≤ 5)P(x ≤ 5) = binomcdf(12, 0.82, 5) = = binomcdf(12, 0.82, 5) = 0.00210.0021

P(x = 3)P(x = 3) = geometpdf(0.82, 3) = = geometpdf(0.82, 3) = 0.02650.0265

P(x ≥ 9)P(x ≥ 9) = 1 – binomcdf(12, 0.82, 8) = = 1 – binomcdf(12, 0.82, 8) = 0.84480.8448= 1 – P(x ≤ 8)= 1 – P(x ≤ 8)

P(x = 5)P(x = 5) 1n kkn

p pk

12 5512

.82 1 .825

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)P(x = 15) = binompdf(30, 0.22, 15) = = binompdf(30, 0.22, 15) = 0.0005110.000511

P(x = 10)P(x = 10) = binompdf(30, 0.22, 10) = = binompdf(30, 0.22, 10) = 0.05540.0554

P(x ≥ 6)P(x ≥ 6) = 1 – binomcdf(30, 0.22, 5) = = 1 – binomcdf(30, 0.22, 5) = 0.67390.6739= 1 – P(x ≤ 5)= 1 – P(x ≤ 5)

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.

μμ = np = 12 · 0.82 = np = 12 · 0.82

σσ = = √√np(1 – p) = np(1 – p) = √√ 12(0.82)(1 – 0.82) 12(0.82)(1 – 0.82)

μμ = = 9.849.84

σσ = = 1.3311.331

Probability from 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 with µ = np and σσ = = √√np(1 – p)np(1 – p) . .

xz

Probability = (__, __)normalcdf

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 Normal Approximation to Binomial Probability

µ = npµ = np µ = 360(.20)µ = 360(.20)

σσ = 7.589 = 7.589σσ = = √√np(1 – p)np(1 – p) σσ = = √√360(.2)(1 – .2)360(.2)(1 – .2)

µ = 72µ = 72

P(x ≤ 50) = P(x ≤ 50) =

Normalcdf(-E99, -2.899) =Normalcdf(-E99, -2.899) =

50 72

7.589z

P(P(zz ≤ ≤ -2.899-2.899) = ) = 0.00190.0019

xz

P(x ≤ 50) = P(x ≤ 50) =

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.

μμ = = npnp

√√ 275(0.95)(0.05)275(0.95)(0.05)σσ = = 3.61423.6142

P(x ≥ 266) = P(x ≥ 266) =

Normalcdf(1.314, E99) =Normalcdf(1.314, E99) =

P(x ≥ 266) = P(x ≥ 266) = 0.09440.0944

µ = µ = 261.25261.25

P(x ≥ 266) = ? P(x ≥ 266) = ?

σσ = = √√np(1 – p) np(1 – p) ==

xz

266 261.25

3.6142z

P(P(zz ≥ ≥ 1.3141.314) = ) =

= 275 · 0.95 = 275 · 0.95

261.

2526

1.25

266

266

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 Geometric Mean µ = 1/pµ = 1/p µ = 1/.20µ = 1/.20 µ = 5µ = 52. What is the probability that the first unbelted driver is the 6th car?

Geometric Prob. Geometric Prob. P(x=6)P(x=6) Geometpdf(.2,6)Geometpdf(.2,6) = 0.0655= 0.06553. What is the probability that the first 10 drivers stopped are wearing

their seatbelts?

Binomial Prob. Binomial Prob. P(x=10)P(x=10) Binompdf(10,.80,10)Binompdf(10,.80,10)= 0.1074= 0.1074

0.80 wear seat belts P(10 out of 10)0.80 wear seat belts P(10 out of 10)

OROR .80 .801010 4. If the police stop 30 cars, how many drivers are expected to be

unbelted? What is the standard deviation?

Binomial Mean Binomial Mean µ = npµ = np µ = 30(.20)µ = 30(.20)σσ = 2.1909 = 2.1909σσ = = √√np(1 – p)np(1 – p) σσ = = √√30(.2)(1 – .2)30(.2)(1 – .2)

µ = 6µ = 6

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. Geometric Prob. P(x≤10)P(x≤10) Geometcdf(.2,10)Geometcdf(.2,10) = 0.8926= 0.8926

Binomial Prob. Binomial Prob. P(x= 4)P(x= 4) Binompdf(10,.20,4)Binompdf(10,.20,4) = .088= .088

7. What is the probability that the police will find at least 4 unbelted drivers within the first 10 drivers stopped?

Binomial Prob. Binomial Prob. P(x≥ 4)P(x≥ 4)1 - Binomcdf(10,.20,3)1 - Binomcdf(10,.20,3) = 0.1209= 0.1209

= 1 – P(x ≤ 3)= 1 – P(x ≤ 3)

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 Normal Approximation to Binomial Probability

µ = npµ = np µ = 250(.50)µ = 250(.50)

σσ = 7.906 = 7.906σσ = = √√np(1 – p)np(1 – p) σσ = = √√250(.5)(1 – .5)250(.5)(1 – .5)

µ = 125µ = 125

P(x ≥ 140) = P(x ≥ 140) =

Normalcdf(1.897, E99) =Normalcdf(1.897, E99) =

140 125

7.906z

P(P(zz ≥ ≥ 1.8971.897) = ) = 0.02890.0289

xz

P(x ≥ 140) = P(x ≥ 140) =

WARM – UP2003 #3

Solution

Part (a):