Discrete Probability distributions

Question 1Determine 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 2Determine 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 x0.4+P(4)+0.1+0.2=1P(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 x0.30+0.15+P(2)+0.20+0.15+0.05=1P(2)=0.15

Question 3In 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 excelSelect 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?

3-7

Remember this from our first class???

(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 activitiesP(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 activitiesP(3U4) = P(3)+P(4)

0.320+0.374=0.694

Question 4Shawn 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 propertyProfit=Sale price-cost priceTotal 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 5According 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 5According 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=10Using the binomial distribution tables:P(10)= 0.2123

Or use formula:

xnxxn ppC )1(

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 5According 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) = 0.4357P(X≥10)= 1-0.4357 = 0.5643

Question 5According 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 tableP(X<10) = P(X≤9) = 0.4357

Question 5According 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 table0.710+0.1319+0.1906+0.2123 = 0.6058

Using the cumulative probability tableP(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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