# Discrete Probability distributions

## Presentation on theme: "Discrete Probability distributions"— Presentation transcript:

Discrete Probability distributions

Question 1 Determine whether the following distribution is a discrete probability distribution. If not state why? x 1 2 3 4 P(x) 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) =1 P(4)=0.3 x 1 2 3 4 5 P(x) 0.30 0.15 ? 0.20 0.05 ∑P(x)=1; for all x P(2) =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) 0.035 1 0.074 2 0.197 3 0.320 4 0.374 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) =0.694

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 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

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. Explain why this is a binomial experiment Find the probability that exactly 10 flights are on time Find the probability that at least 10 flights are on time Find the probability that fewer than 10 flights are on time Find the probability that between 7 and 10 flights inclusive are on time Solution: This is a binomial distribution because it satisfies each of the four requirements: There are fixed number of trials (n=15) The trials are all independent (randomly selected) For each trial, there are only two possible outcomes(“on-time” and “not on time”) 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. 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)= Or use formula:

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. 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: P(X≥10) = 1-P(X<10) = 1- P(X≤9) Using cumulative binomial table, P(X≤9) = P(X≥10)= =

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) =

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 = Using the cumulative probability table P(7≤ X ≥10) = P(X≤10)-P(X≤6) = =