Transcript Chapter 6.1: Probability distributions

Discrete Probability
distributions
Question 1
Determine whether the following distribution is a discrete probability distribution.
If not state why?
x
0
1
2
3
4
P(x)
0.2
0.2
0.2
0.2
0.2
Yes, because ∑ P(x)=1 and 0≤P(x)≤1 for all x
x
10
20
30
40
50
P(x)
0.1
0.23
0.22
0.6
-0.15
No, because ∑ P(50)<0
Question 2
Determine the required value of the missing probability to make the distribution
a discrete probability distribution:
x
3
4
5
6
P(x)
0.4
?
0.1
0.2
∑P(x)=1; for all x
0.4+P(4)+0.1+0.2=1
P(4)=0.3
x
0
1
2
3
4
5
P(x)
0.30
0.15
?
0.20
0.15
0.05
∑P(x)=1; for all x
0.30+0.15+P(2)+0.20+0.15+0.05=1
P(2)=0.15
Question 3
In the following probability distribution, the random variable X represents the
number of activities a parent of a K-5th grade student is involved.
X
P(x)
0
0.035
1
0.074
2
0.197
3
0.320
4
0.374
(a) Verify that this is a discrete probability distribution.
This is a discrete probability distribution because all the probabilities are
between 0 and 1(inclusive) and the sum of the probabilities is 1.
(b) Draw a probability histogram.
Steps:
Select data series in excel
Select Insert tab, click on column and select type of chart.
What do you think the distribution shape is?
What do you think will be the relationship between the mean and the median?
Remember this from our first class???
3-7
(c) Compute and interpret the mean of the random variable X.
(d) Compute the variance of the random variable X
(e) Compute the Standard deviation of the random variable X
(f) What is the probability that a randomly selected student has a parent
involved in 3 activities?
Solution:
P(3)= The probability that a student has a parent involved in 3 activities
P(3) = 0.320
(g) What is the probability that a randomly selected student has a parent
involved in 3 or 4 activities?
Solution:
P(3U4)= The probability that a student has a parent involved in 3 or 4 activities
P(3U4) = P(3)+P(4)
0.320+0.374
=0.694
Question 4
Shawn and Maddie purchase a foreclosed property for $50,000 and spend an
additional $27,000 fixing up the property. They feel that they can resell the
property for $120,000 with probability 0.15, $100,000 with probability 0.45,
$80,000 with probability 0.25 and $60,000 with probability 0.15. Compute and
interpret the expected profit from re-selling the property.
Solution:
Let x= the profit for re-selling the property
Profit=Sale price-cost price
Total cost of purchasing and fixing property is $50,000 + $27,000 = $77,000
Selling Price ($)
120,000
100,000
80,000
60,000
Profit, x ($)
43,000
23,000
3,000
-17,000
Probability
0.15
0.45
0.25
0.15
Shawn and Maddie can expect to earn a profit of $15,000 on the average, if
they resold the property.
Binomial Probability
Question 5
According to flightstats.com, American Airlines flight 1247 from Orlando to Los
Angeles is on time 65% of the time. Suppose fifteen flights are randomly
selected, and the number of on-time flights is recorded.
(a) Explain why this is a binomial experiment
(b) Find the probability that exactly 10 flights are on time
(c) Find the probability that at least 10 flights are on time
(d) Find the probability that fewer than 10 flights are on time
(e) Find the probability that between 7 and 10 flights inclusive are on time
Solution:
(a) This is a binomial distribution because it satisfies each of the four
requirements:
(1) There are fixed number of trials (n=15)
(2) The trials are all independent (randomly selected)
(3) For each trial, there are only two possible outcomes(“on-time” and “not on
time”)
(4) The probability of successes(i.e “on-time”) is the same for all trials (0.65)
Question 5
According to flightstats.com, American Airlines flight 1247 from Orlando to Los
Angeles is on time 65% of the time. Suppose fifteen flights are randomly
selected, and the number of on-time flights is recorded.
(a) Find the probability that exactly 10 flights are on time
Solution:
n=15, p=0.65 and x=10
Using the binomial distribution tables:
P(10)= 0.2123
Or use formula:
C x p (1  p)
x
n
n x
Using excel to find the binomial distribution
According to flightstats.com, American Airlines flight 1247 from Orlando to Los
Angeles is on time 65% of the time. Suppose fifteen flights are randomly selected,
and the number of on-time flights is recorded.
(a) Find the probability that exactly 10 flights are on time
Question 5
According to flightstats.com, American Airlines flight 1247 from Orlando to Los
Angeles is on time 65% of the time. Suppose fifteen flights are randomly
selected, and the number of on-time flights is recorded.
(c) Find the probability that at least 10 flights are on time
Solution:
(a) P(X≥10) = 1-P(X<10) = 1- P(X≤9)
Using cumulative binomial table, P(X≤9) = 0.4357
P(X≥10)= 1-0.4357 = 0.5643
Question 5
According to flightstats.com, American Airlines flight 1247 from Orlando to Los
Angeles is on time 65% of the time. Suppose fifteen flights are randomly
selected, and the number of on-time flights is recorded.
(d)Find the probability that fewer than 10 flights are on time
Solution:
Using cumulative binomial table
P(X<10) = P(X≤9) = 0.4357
Question 5
According to flightstats.com, American Airlines flight 1247 from Orlando to Los
Angeles is on time 65% of the time. Suppose fifteen flights are randomly
selected, and the number of on-time flights is recorded.
(e)Find the probability that between 7 and 10 flights inclusive are on time
Solution:
P(7≤ X ≥10) = P(7)+P(8)+P(9)+P(10)
Using the binomial probability table
0.710+0.1319+0.1906+0.2123 = 0.6058
Using the cumulative probability table
P(7≤ X ≥10) = P(X≤10)-P(X≤6) = 0.6481-0.0422 = 0.6059
CSTEM Web link
http://www.cis.famu.edu/~cdellor/math/
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