5-3 Homework: Chapter 6 (Sections 6.1 through 6.4)

A lottery ticket has a grand prize of $45.2 million. The probability of winning the grand prize is .000000044. Determine the expected value of the lottery ticket. Is paying $1 for a ticket greater than the expected winnings?

1. 45,200,000 multiplied by 0.000000044 = 1.989 2. YES Explanation: X is the lottery ticket winnings. The values of X are either $45.2 million or $0. The expected winning is E(X) = $45,200,000(.000000044) + $0(.999999956) = $1.989. Because the lottery ticket costs less than the expected value ($1.00 < $1.989), you should buy the ticket.

Oxnard Petro Limited is buying hurricane insurance for its off-coast oil drilling platform. During the next five years, the probability of total loss of only the above-water superstructure ($320 million) is .30, the probability of total loss of the facility ($820 million) is .30, and the probability of no loss is .40. Find the expected loss: _______ million

342 Step-by-step explanation: The formula to compute the expected loss is: E(Loss) = ∑Loss amount • P(Loss) Given: P (Loss) Amount of loss 0.30 $320 million 0.30 $820 million 0.40 $0 Compute the expected loss as follows: E(Loss) = ∑Loss amount • P(Loss). = (0.30•320) + (0.30•820) + (0.4•0) in millions = 96 + 246 = 0 = $342 Million Thus, the expected loss is $342 million. FURTHER Explanation: Let X equal the loss during a hurricane. The values of X are either $320 million, $820 million, or $0. Expected Loss = E(X ) = $320(.30) + $820(.30) + $0(.40) = $342 million.

On the midnight shift, the number of patients with head trauma in an emergency room has the probability distribution shown below. x: 0 1 2 3 4 5 Total P(x) .05 .34 .26 .20 .10 .05 1.00 (a) Calculate the mean and standard deviation. (b) Describe the shape of this distribution.

A) MEAN: μ = 2.11 math: In order to find the mean (expectation) of the given distribution, we will use the following formula: μ = ∑x • P(x) So, first we need to multiply each value of X by each probability P(X), then add these results together. In this example we have: μ = 0 multiplied by .05 = 0 1 multiplied by .34 = 0.34 2 multiplied by .26 = 0.52 3 multiplied by .20 = 0.60 4 multiplied by .10 = 0.40 5 multiplied by .05 = 0.25 TOTAL = 2.11 STANDARD DEVIATION: σ = 1.2561 math: In order to find the standard deviation of the given distribution, we will use the following formula: σ = √∑x²•P(x)-µ² In this example: µ = 2.11 Now we will find the sum: ∑x²•P(x) = 0² • 0.05+1² • 0.34+2² • 0.26+3² • 0.2+4² • 0.1+5² • 0.05 ∑x²•P(x) = 6.03 Putting all together we have: σ=√∑x² • p(x) − μ² = √6.03−2.11² =√1.578 ≈ 1.2561 B) Skewed to the right

On hot, sunny, summer days, Jane rents inner tubes by the river that runs through her town. Based on her past experience, she has assigned the following probability distribution to the number of tubes she will rent on a randomly selected day. X: 25 50 75 100 TOTAL P(x): .27 .30 .32 .11 1.00 (a) Find the probability expressions. a.P(X = 75) b.P(X ≤ 50) c.P(X > 25) d.P(X < 75) (b) Which of the probability expressions in parts (a) to (d) is a value of the CDF?

A) a: 0.32 math: P(X = 75) = .32. b: 0.57 math: the values in x should be less than or equal to 50, which are number's 25 & 50. The corresponding P(x) values are .27 & .30. Add both and get 0.57. c: 0.73 math: add all P(x) values that have an x value greater than 25, which are .30, .32, .11. d: 0.57 math: add all P(x) values that has an x value less than 75, which are x values 25 & 50. Adding the corresponding P(x) values would be .27 & .30 = .57 B) Part B is a CDF because a CDF is defined by the inequality "≤."

Find the mean and standard deviation for each binomial random variable: a. n = 35, π = .70 b. n = 74, π = .50 c. n = 25, π = .85

A: mean: 24.50 standard deviation: 2.7111 B: mean: 37.00 standard deviation: 4.3012 C: mean: 21.25 standard deviation: 1.7854 Explanation a. μ = (35)(.70) = 24.50. σ = √(35)(.70)(1−.70)=2.7111 b. μ = (74)(.50) = 37.00. σ = √(74)(.50)(1−.50)=4.3012 c. μ = (25)(.85) = 21.25. σ = √(25)(.85)(1−.85)=1.7854

Which of the following could not be probability distributions? Example A x P(x) 0 .40 1 .90 Example B x P(x) 1 .10 2 .20 3 .30 4 .20 5 .10 Example C x P(x) 50 .30 60 .60 70 .10

EXAMPLE A & B EXPLANATION: Example C is a probability distribution because the sum of P(x) is 1 (0.3 + 0.6 + 0.1 = 1) and all probabilities are nonnegative. Example A and B are not probability distributions because the sum of P(x) is 1.30 for A and .90 for B.

The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random variable ranging from 4 to 10. Find the mean and standard deviation of this random variable.

MEAN: 4+10=14 14/2 = 7 MEAN = 7 STANDARD DEVIATION: 2.00 First, work out the average, or arithmetic mean, of the numbers: Count: 7 (How many numbers) Sum: 49 (All the numbers added up) Mean: 7 (Arithmetic mean = Sum / Count) Then, take each number, subtract the mean and square the result: Differences: (Every Number minus Mean) 4-7 = -3 5-7 = -2 6-7 = -1 7-7 = 0 8-7 = 1 9-7 = 2 10-7 = 3 Differences²: (Square of each difference) -3² = 9 -2² = 4 -1² = 1 0² = 0 1² = 1 2² = 4 3² = 9 Now calculate the Variance: Sum of Differences²: (Add up the Squared Differences) = 28 Variance: (Sum of Differences² / Count) = 4 Lastly, take the square root of the Variance: √4 = 2 Standard Deviation: 2.00 FURTHER Explanation Using a = 4 and b = 10, μ = (4 + 10)/2 = 7.00. σ=√[(10−4+1)²−1]/12 = 2.00

At a Noodles & Company restaurant, the probability that a customer will order a nonalcoholic beverage is .32. a. Find the probability that in a sample of 7 customers, none of the 7 will order a nonalcoholic beverage. b. Find the probability that in a sample of 7 customers, at least 4 will order a nonalcoholic beverage. c. Find the probability that in a sample of 7 customers, fewer than 5 will order a nonalcoholic beverage. d. Find the probability that in a sample of 7 customers, all 7 will order a nonalcoholic beverage.

a: 0.0672 math: the probability of not ordering a nonalcoholic beverage is 0.68 (1 - 0.32). Multiply 0.68 to the 7th power = 0.06722988818432 b: 0.1534 c: 0.9620 math: find the probability of numbers 0-4 and add the sum: P(0) = 0.0672 P(1) = 0.2215 P(2) = 0.3127 P(3) = 0.2452 P(4) = 0.1154 d: 0.32ˆ8 = 0.0003 Explanation: The number of customers out of the next 7 who order a nonalcoholic beverage is a binomial random variable with n = 7 and π = .32. a. P(X = 0) = BINOM.DIST(0,7,.32,0) = .0672 b. P(X ≥ 4) = 1 − P(X ≤ 3) = 1 −BINOM.DIST(3,7,.32,1) = .1534 c. P(X < 5) = P(X ≤ 4) =BINOM.DIST(4,7,.32,1) = .9620 d. P(X = 7) = BINOM.DIST(7,7,.32,0) = .0003

Write the probability of each italicized event in symbols (e.g., P(X ≥ 5)). (a) At least 4 correct answers on a 15-question quiz (X = number of correct answers). (b) Fewer than 5 "phishing" e-mails out of 25 e-mails (X = number of phishing e-mails). (c) At most 5 no-shows at a party where 16 guests were invited (X = number of no-shows).

a: P(X ≥ 4) b: P(X < 5) c: P(X ≤ 5)