Transcript File - phsapstatistics

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.
n k
12 
nk
12 5
P(x = 5)     p 1  p      .825 1  .82 
k 
5
P(x = 5) = binompdf(12, 0.82, 5) = 0.0018
2.
The prob. of making at most 5 baskets.
3.
P(x ≤ 5) = binomcdf(12, 0.82, 5) = 0.0021
The prob. of making at least 9 baskets.
P(x ≥ 9) = 1 – P(x ≤ 8) = 1 – binomcdf(12, 0.82, 8) = 0.8448
4. The prob. he will make his first basket on the 3rd throw.
P(x = 3) = geometpdf(0.82, 3) = 0.0265
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?
P(x = 15) = binompdf(30, 0.22, 15) = 0.000511
(b)…a third of the class get 5’s?
P(x = 10) = binompdf(30, 0.22, 10) = 0.0554
(c)…at least 6 students get 5’s?
P(x ≥ 6) = 1 – P(x ≤ 5) = 1 – binomcdf(30, 0.22, 5) = 0.6739
Binomial Mean and Standard Deviation
A Binomial Distribution, B(n, p) has:
Mean =
Standard Deviation =
μ = np
σ = √np(1 – p)
A college basketball player has an 82% probability
of making Free throws. During a typical game he usually
attempts 12 free throws. Find:
EXAMPLE:
1. The mean number of baskets he is expected to make.
μ = np = 12 · 0.82
μ = 9.84
2. The Standard Deviation of the number of baskets.
σ = √np(1 – p) = √ 12(0.82)(1 – 0.82) σ = 1.331
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 σ = √np(1 – p) .
z
x

Probability = normalcdf (__, __)
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.
EXAMPLE:
Normal Approximation to Binomial Probability
P(x ≤ 50) =
z
µ = np µ = 360(.20) µ = 72
σ = √np(1 – p) σ = √360(.2)(1 – .2) σ = 7.589
50  72
z
P(x ≤ 50) =
7.589
P(z ≤ -2.899) = Normalcdf(-E99, -2.899) = 0.0019
x

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) = ?
EXAMPLE:
z
x

μ = np = 275 · 0.95
σ = √np(1 – p) =
266  261.25
z

P(x ≥ 266) =
3.6142
P(z ≥ 1.314) = Normalcdf(1.314, E99) =
P(x ≥ 266) = 0.0944
µ = 261.25
√ 275(0.95)(0.05)
σ = 3.6142
Police estimate that 20% of drivers do not wear
their seatbelts. They set up a Safety Roadblock to check
for seatbelt usage.
EXAMPLE:
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 6th 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? 0.80 wear seat belts P(10 out of 10)
Binomial Prob.
P(x=10) Binompdf(10,.80,10)
OR
.8010 = 0.1074
4. If the police stop 30 cars, how many drivers are expected to be
unbelted? What is the standard deviation?
µ = np µ = 30(.20)
Binomial Mean
σ = √np(1 – p) σ = √30(.2)(1 – .2)
µ=6
σ = 2.1909
Police estimate that 20% of drivers do not wear
their seatbelts. They set up a Safety Roadblock to check
for seatbelt usage.
EXAMPLE:
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 the first unbelted driver
within the first 10 drivers stopped
Geometric Prob. P(x≤10) Geometcdf(.2,10) = 0.8926
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 – P(x ≤ 3)
1 - Binomcdf(10,.20,3)
= 0.1209
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?
EXAMPLE:
Normal Approximation to Binomial Probability
P(x ≥ 140) =
z
µ = np µ = 250(.50) µ = 125
σ = √np(1 – p) σ = √250(.5)(1 – .5) σ = 7.906
140  125
P(x ≥ 140) = z 
7.906
P(z ≥ 1.897) = Normalcdf(1.897, E99) = 0.0289
x

WARM – UP
2003 #3
Solution
Part (a):