(a)
a. 0.30
b. 0.20+0.40+0.30=0.90
c. 0.30+0.10=0.40
d. 0.20+0.40+0.30=0.90
(b)
Part (b)
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.
(a) Find the
Prize*Probability of Getting Prize
Expected Value=215,000 * 0.00000884=1.9006
Pepsi and Mountain Dew products sponsored a contest giving away a Lamborghini sports car worth $215,000. The probability of winning from a single bottle purchase was .00000884.
Find the expected value. (Round your answer to 4 decimal places.)
$28 million * .000000023 = .644
A lottery ticket has a grand prize of $28 million. The probability of winning the grand prize is .000000023.
Determine the expected value of the lottery ticket. (Round your answer to 3 decimal places.)
(a) x >/= 7
(b) x<4
(c) x</= 2
Write the probability of each italicized event in symbols (e.g., P(X ? 5).
(a) At least 7 correct answers on a 10-question quiz (X = number of correct answers).
(b) Fewer than 4 "phishing" e-mails out of 20 e-mails (X = number of phishing e-mails).
(c) At
(a)
Excel: =BINOM.DIST(2,8,0.1,0)
P(X = 2): 0.1488
(b)
Excel: =BINOM.DIST(1,10,0.4,0)
P(X = 1): 0.0403
(c)
Excel: =BINOM.DIST(3,12,0.7,0)
P(X = 3): .0014
Calculate each binomial probability:
(a) X = 2, n = 8, ? = .10 (Round your answer to 4 decimal places.)
P(X = 2)
(b) X = 1, n = 10, ? = .40 (Round your answer to 4 decimal places.)
P(X = 1)
(c) X = 3, n = 12, ? = .70 (Round your answer to 4 decimal places
(a) =Binom.Dist(3,12,.01,1)
Probability: 0.9744
(b)=1-BINOM.DIST(2,7,0.4,1)
Probability: 0.5801
(c) =Binom.Dist(9,14,0.6,1)
Probability: .7207
Calculate each binomial probability:
(a) Fewer than 4 successes in 12 trials with a 10 percent chance of success. (Round your answer to 4 decimal places.)
(b) At least 3 successes in 7 trials with a 40 percent chance of success. (Round your answer to 4 de
(a) =BINOM.DIST(0,12,0.05,0)
Probability: 0.5404
(b) =BINOM.DIST(1,12,0.05,0)
Probability: 0.3413
(c) =BINOM.DIST(2,12,0.05,0)
Probability: 0.0988
(d) =BINOM.DIST(2,12,0.05,1)
Probability: 0.9804
(e-1) Graph A
(e-2) Skewed to the right
Historically, 5 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database.
The number of customers out of 12 who have an incorrect address in the database is a binomial random variable
(a) =BINOM.DIST(8,8,0.6,0)
Probability: .0168
(b) =1-BINOM.DIST(4,8,0.6,1)
Probability: .5941
(c) =1-BINOM.DIST(4,8,0.6,1)
Probability: .5941
(d) Mean: 8 x .6 = 4.80
Standard Deviation: =SQRT(8
.6
.4)
= 1.3856
(e-1) Graph A
(e-2) Skewed left
J.D. Power and Associates says that 60 percent of car buyers now use the Internet for research and price comparisons.
(a) Find the probability that in a sample of 8 car buyers, all 8 will use the Internet. (Round your answer to 4 decimal places.)
(b) Find
(a) =BINOM.DIST(0,12,0.15,0)
P(X=0): 0.1422
(b) =BINOM.DIST(1,12,0.15,0)
P(X=1): 0.3012
(c) =1-BINOM.DIST(1,12,0.15,1)
P(X ? 2): 0.5565
Police records in the town of Saratoga show that 15 percent of the drivers stopped for speeding have invalid licenses. If 12 drivers are stopped for speeding.
(a) Find the probability that none will have an invalid license. (Round your answer to 4 decimal
(a) Mean
79.6 x 2.54 = 202.184
(b) Standard Deviation
3.24 x 2.54 = 8.2296
(c) Rule 1 for the MEAN and Rule 2 for the STANDARD DEVIATION
The height of a Los Angeles Lakers basketball player averages 6 feet 7.6 inches (i.e., 79.6 inches) with a standard deviation of 3.24 inches. To convert from inches to centimeters, we multiply by 2.54.
(a) In centimeters, what is the mean? (Round your ans
(a) ? = 37.1�F
? = (5/9)(37.1) - 17.78 = 2.83� C
(b) ? = 10.3 �F
? = (5/9)(10.3) = 5.72� C.
The mean January temperature in Fort Collins, CO, is 37.1� F with a standard deviation of 10.3� F.
Express these Fahrenheit parameters in degrees Celsius using the transformation C = (5/9)F ? 17.78. (Round your answers to 2 decimal places.)