Transcript Chapter 6 Review

Lesson 6 - R
Review of Chapter 6
Discrete PDFs
Binomial and Geometeric
Objectives
• Explain what is meant by a binomial setting and
binomial distribution.
• Use technology to solve probability questions in a
binomial setting.
• Calculate the mean and variance of a binomial
random variable.
• Solve a binomial probability problem using a Normal
approximation.
• Explain what is meant by a geometric setting.
• Solve probability questions in a geometric setting.
• Calculate the mean and variance of a geometric
random variable.
Vocabulary
• None new
Using your TI-83 calculator
We can use 1-Var-Stats to calculate the mean and
standard deviation of a discrete random variable given
it’s outcomes and probability
• Type in outcomes in L1
• Type in corresponding probabilities in L2
• Use 1-Var-Stats L1, L2 to get statistics
• Notes:
– Discrete Random Variables have countable (finite) values
– Continuous Random Variables have an interval of values
(infinite)
– Ranges of Random Variables are determined by minimum or
maximum values that they can take on
Discrete Random Variable - Mean
The mean, or expected value [E(x)], of a discrete
random variable is given by the formula
μx = ∑ [x ∙P(x)]
where x is the value of the random variable and
P(x) is the probability of observing x (multiply them
together and add all of them up)
Mean of a Discrete Random Variable Interpretation:
If we run an experiment over and over again, the
law of large numbers helps us conclude that the
difference between x and ux gets closer to 0 as n
(number of repetitions) increases
Discrete Random Variable - Variance
Variance and Standard Deviation of a Discrete RV:
The variance of a discrete random variable is given by:
σ2x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x2 ∙ P(x)] – μ2x
and standard deviation is √σ2
Note: round the mean, variance and standard deviation
to one more decimal place than the values of the
random variable
Means and Variances
• Rules for Means
– Means follow the rules for linear combinations (from Algebra)
– When you linearly combine two or more (rules give only the 2
case example) random variables, you combine their means in
the same manner
• E(a + X + bY) = a + E(X) + bE(Y)
• Rules for Variances
– Adding a number to a random variable does not change its
variance
– Multiply a random variable by a number changes the variance
by the square of that number
• V(a + X + bY) = V(X) + b²V(Y)
– When you combine random variables, you always add the
variances
• V(X - Y) = V(X) + V(Y) = V(X + Y)
English Phrases
Math
Symbol
≥
>
<
≤
=
≠
English Phrases
At least
More than
Fewer than
No more than
Exactly
Different from
No less than
Greater than
Less than
At most
Equals
P(x ≤ A) = cdf (A)
Greater than or equal to
Less than or equal to
Is
P(x = A) = pdf (A)
P(X)
∑P(x) = 1
Cumulative
probability
or cdf
P(x ≤ A)
Values of Discrete Variable, X
P(x > A) = 1 – P(x ≤ A)
X=A
Binomial Probability Criteria
A random variable is said to be a binomial provided:
1. For each trial there are two mutually exclusive
(disjoint) outcomes: success or failure
2. The trials are independent
3. The probability of success is the same for each trial of
the experiment
4. The experiment is performed a fixed number of times.
Each repetition is called a trial
Most important skill for using binomial distributions is
the ability to recognize situations to which they do and
don’t apply
Binomial PDF
The probability of obtaining x successes in n
independent trials of a binomial experiment, where the
probability of success is p, is given by:
P(x) = nCx px (1 – p)n-x,
nCx
x = 0, 1, 2, 3, …, n
is also called a binomial coefficient and is defined by
combination of n items taken x at a time or
where n! is n  (n-1)  (n-2)  …  2  1
n
k
n!
= -------------k! (n – k)!
Geometric Probability Criteria
A random variable is said to be a geometric provided:
1. For each trial there are two mutually exclusive (disjoint)
outcomes: success or failure
2. The trials are independent
3. The probability of success is the same for each trial of
the experiment
4. We repeat the trials until we get a success
Geometric PDF
When we studied the Binomial distribution, we were
only interested in the probability for a success or a
failure to happen. The geometric distribution
addresses the number of trials necessary before the
first success. If the trials are repeated k times until the
first success, we will have had k – 1 failures. If p is the
probability for a success and q (1 – p) the probability
for a failure, the probability for the first success to
occur at the kth trial will be (where x = k)
P(x) = p(1 – p)x-1,
x = 1, 2, 3, …
The probability that more than n trials are needed
before the first success will be
P(k > n) = qn = (1 – p)n
Means and Normal Apx to Binomial
• Means and Standard Deviations
– Binomial
• Mean: E(X) = μ = np
• Variance: ² = np(1 – p)
– Geometric
• Mean: E(X) = μ =1/p
• Variance: ² = (1- p) / p²
• Normal distribution N(μ,σ) can approximate a
Binomial curve, when conditions are met
1. n < 0.10N (sample small enough – independence)
2. np ≥ 10 and n(1-p) ≥ 10 (for normality)
TI-83 Reminders
• Binomial
– N: number of trials
– P: probability of success
– X: number of successes
• Geometric
– P: probability of success
– X: number of trials until first success
• Remember to use catalog help
• PDF X = #
• CDF  X ≤ #
• Complement Rule for X ≥ #
TI-83 Binomial Support
• For P(X = k) using the calculator: 2nd VARS
binompdf(n,p,k)
• For P(k ≤ X) using the calculator: 2nd VARS
binomcdf(n,p,k)
• For P(X ≥ k) use 1 – P(k < X) = 1 – P(k-1 ≤ X)
TI-83 Geometric Support
• For P(X = k) using the calculator: 2nd VARS
geometpdf(p,k)
• For P(k ≤ X) using the calculator: 2nd VARS
geometcdf(p,k)
• For P(X > k) use 1 – P(k ≤ X) or (1- p)k
Non AP Distributions - ID
• Hypergeometric
– Small population sampling without replacement
– Example: drawing names out of a hat
• Negative Binomial
– Number of trials until the nth success
– Example: number of foul shots until his 3rd successful one
– Geometric is a special case of this (n = 1)
• Poisson
– Successes spread over spatial random variable
(time or area)
– Example: arrivals per minute at McD, potholes per mile on I-81
Example 1a/b: Which PDF?
Determine which probability distribution (Binomial,
Negative Binomial, Geometric, Hyper-geometric, and
Poisson) best fits the following. Use only once.
a. A stats class using a bucket filled with 20 red and
20 green balls, pulls a ball out of the bucket and
records its color. They record the number of pulls
required until they have 5 green balls pulled and
repeat whole process 50 times.
b. A stats class using a bucket filled with 20 red and
20 green balls, drops the bucket and scatters the
balls across the room. They record the number of
balls per floor tile and repeat this 50 times.
Which PDF?
Determine which probability distribution (Binomial,
Negative Binomial, Geometric, Hyper-geometric, and
Poisson) best fits the following. Use only once.
c. A stats class using a bucket filled with 20 red and
20 green balls, pulls a ball out of the bucket, records
its color, replaces it and repeats until they pull a
green ball. They record the number of pulls before
the green ball is pulled out.
d. A stats class using a bucket filled with 20 red and
20 green balls, pulls a ball out of the bucket and
records its color and repeat it 50 times
Which PDF?
Determine which probability distribution (Binomial,
Negative Binomial, Geometric, Hyper-geometric, and
Poisson) best fits the following. Use only once.
e. A stats class using a bucket filled with 20 red and
20 green balls, pulls 5 balls out of the bucket and
records the number of red balls and repeat it 50
times.
a.
b.
c.
d.
e.
Negative Binomial (pull until rth success)
Poisson (successes over an area)
Geometric (pulls till first success)
Binomial (with n=1)
Hyper-geometric (w/o replacement)
Summary and Homework
• Summary
– Use pdf for an X = #
– Use cdf for an X ≤ #
– Use complement rule for X ≥ #
• P(X ≥ #) = 1 – P(X < #)
• P(X > #) = 1 – P(
– Binomial – Bernoulli with fixed # of trials
• Mean: np
Variance: np(1-p)
– Geometric – Bernoulli until first success
• Mean: 1/p
• Homework:
Variance: (1-p)/p²
Problem 1a
The random variable X represents the number of people
that you have to wait behind in line when you go to the
post office to buy stamps at lunch time. The probability
distribution of X is provided below:
X=
Probability =
0
.1
1
.5
2
.3
3
.1
(a) Find the mean number of people that will be in front of
you in the stamp line. Use the definition and show
work.
Mean: ∑ [x ∙P(x)] = (.1)(0) + (.5)(1) + (.3)(2) + (.1)(3)
= 0 + .5 + .6 + .3 = 1.4
Problem 1b
The random variable X represents the number of people
that you have to wait behind in line when you go to the
post office to buy stamps at lunch time. The probability
distribution of X is provided below:
X=
Probability =
0
.1
1
.5
2
.3
3
.1
(b) Find the standard deviation for the number of people
in the line in front of you. Use the definition and show
work.
Var: ∑[x2 ∙ P(x)] – μ2x = ∑ [x2 ∙ P(x)] – μx2
= (0 + .5 + .3(4) + .1(9) ) – 1.96)
= 2.6 – 1.96
= 0.64
St Dev
= 0.8
Problem 2
From the previous problem, let f(X) = 2X + 0.5 represent
the amount of time (in minutes) required for the clerks to
process X people. Show your work and use the shortcut
methods (not the definitions) to find:
(i) The mean number of minutes that you will have to
wait.
μX = 1.4
so μf(X) = 0.5 + 2 μX = 0.5 + 2(1.4) = 3.3 minutes
(ii) The standard deviation of the number of minutes you
will have to wait.
σX = 0.8
so σ²f(X) = 2² σ²X = 4 (0.8)² = 2.56 minutes
σf(X) = 2.56 = 1.6 minutes
Problem 3
While you are at the post office you also need to pick up a
package. The random variable Y represents the number
of people you have to wait behind in the pickup line. The
probability distribution of Y is provided below:
Y=
Probability =
0
.2
1
.3
2
.5
(d) Use your calculator to find the mean number of people
that will be in front of you in this line
Mean: ∑ [x ∙P(x)] = 1.3
(e) Use your calculator to find the standard deviation of
the number of people in this line
Var: ∑[x2 ∙ P(x)] – μ2x = 0.781
Problem 4
Suppose that the numbers of people in the two lines are
independent of each other. Let Z = X + Y represent the
total number of people you will have to wait behind at the
post office. Use the rules we discussed in class to find:
(i) The mean or expected value of Z.
E(Z) = E(X) + E(Y) = 1.4 + 1.3 = 2.7
(ii) The standard deviation of Z.
V(Z) = V(X) + V(Y) = 0.8² + 0.781² = 1.25
σ(Z) = 1.25 = 1.118
Problem 5
The weight of eggs produced by a certain breed of hen
is normally distributed with mean of μ = 65 grams and
standard deviation of σ = 5 grams. What is the
probability that the weight of a dozen (12) randomly
selected eggs is between 750 grams and 800 grams?
E(D) = μD = μE1 + μE2 + … + μE12 = 12  μE
= 12  65 = 780 grams
V(D) = σ²D = σ²E1 + σ²E2 + σ²E3 … + σ²E12 = 12  σ²E
= 12  5² = 300 grams
σD = 300 = 17.32
normalcdf(750, 800, 780, 17.32) = 83.43%
750
780
800
Problem 6
(a) The binomial setting and the geometric setting are
similar in that they both involve
1) success or failure (mutually exclusive or binary outcomes)
2) probability of success is constant
3) independent outcomes (trials)
(b) How do the binomial and geometric settings differ?
Binomial
fixed number of trials
Geometric
repeat trials until first success
Problem 7
According to the manufacturers, 13% of the M&M’s produced today
are brown. (Did you know that at one time all M&M’s were brown?)
Assume that all large bags of M&M’s contain 13% brown. Suppose
you start taking individual candies out of a large bag, hoping for a
brown one. Let X represent the number of the draw on which you
get your first brown M&M.
(a) On average, how many M&M’s would you expect to select in
order to find a brown one?
X~G(0.13)
E(X) = 1/p = 1/(0.13) = 7.69
(b) Construct a table showing the probability distribution for X (up
through X = 5). Show work for probabilities in the space below.
Round probabilities to 3 decimal places.
X=
Probability =
1
0.130
2
0.113
3
0.098
4
0.086
5
0.074
Problem 7 cont
According to the manufacturers, 13% of the M&M’s produced today
are brown. (Did you know that at one time all M&M’s were brown?)
Assume that all large bags of M&M’s contain 13% brown. Suppose
you start taking individual candies out of a large bag, hoping for a
brown one. Let X represent the number of the draw on which you
get your first brown M&M.
1
2
3
4
5
0.130 0.113
0.098 0.086 0.074
(c) Construct a histogram that shows the cumulative probability
distribution for X (up through X = 5). Label the height of each
bar in addition to providing a scale on the vertical axis.
0.5
0.51
0.4
0.44
0.35
Probability 0.3
0.2
0.1
0.24
0.13
1
2
3
4
5
Nr of trials until first brown M&M
Problem 8
When an oil company conducts exploratory oil drilling, each well is
classified as a producer well or a dry well. Past experience shows
that 15% of all wells drilled are producer wells. The company has
plans to drill at 12 new locations.
(a) What is the probability that exactly three wells will be producer
wells? Be sure to provide support for your answer.
X ~ B(0.15,12)
P(X=3) = 0.1720
binompdf(12,.15,3)
(b) Calculate the probability that at least three wells will be
producer wells. Be sure to provide support for your answer.
X ~ B(0.15,12)
P(X≥3) = 1 – P(X<3) = 1 – P(X ≤ 2) = 0.2642
1 - binomcdf(12,.15,2)
Problem 9
A seed producer claims that 95% of a certain type seed will
germinate under ideal conditions. A testing agency attempts to
germinate 3000 of these seeds.
X ~ B(0.95,3000)
(a) Give the mean and standard deviation for the number of seeds
that would germinate if the producer’s claim is correct.
Mean = np = .95(3000) = 2850
Standard deviation = √np(1-p)
√3000(.95(.05) = 11.94
(b) 2830 of the testing agency’s seeds eventually germinate. Use a
normal approximation to estimate the probability that 2830 or
fewer seeds would germinate if the producer’s claim is correct.
Show work. Check conditions: assume > 30000 seeds produced
np ≥ 10  n(1-p) ≥ 10 
2850 ≥ 10
150 ≥ 10
X ~ N(2850,11.94)
P(X≤2830) = 0.047
normalcdf(-E99,2830,2850,11.94)