6.1
Make a histogram of the following probability distribution:
X= 0 1 2 3 4 5
P(x)= 0.11 0.18 0.21 0.15 0.17 0.09
0 bar to 0.11
1 bar to 0.18
2 bar to 0.21
3 bar to 0.15
4 bar to 0.17
5 bar to 0.09
If X and Y are independent random variables, then how do the sum and difference of the variance of two independent random variables look like?
σ 2x+y=σ 2x+σ 2y
What are the four conditions for the binomial setting?
What's Binary, Independent, Number, Success
Which of the following is geometric?
(a) The number of times I have to flip a coin to get three heads.
(b) The number of dice rolls I make until I roll a four.
(c) The total number of heads I get if I flip a coin 20 times.
(d) The number of 9s in a random row of digits.
(e) The number of 2s in a row of 50 random digits.
B
A casino offers a game where a player pays $10 to roll a fair six-sided die. The player wins according to the following payout structure:
Define the random variable X as the net profit for a player (winnings minus the $10 cost to play). Construct the probability distribution for X.
X is the number of mice caught in traps during a single night in a small apartment building
X= 0 1 2 3 4 5
P(X)= 0.11 0.18 0.21 0.15 0.17 0.09
Describe P(X is > or = to 2) in words and find its value
The probability of having more than or equal to two mice caught in a single night in a small apartment building is 0.91.
0.11+0.18+0.21+0.15+0.17+0.09= 0.91
A _____________ transformation of a random variable involves adding or subtracting a constant a, multiplying or dividing by a constant b, or both
linear
The count X of successes in a binomial setting is a ________________. The probability distribution of X is a __________________ with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial.
Binomial Random Variable, Binomial Distribution
A certain vending machine offers 12-ounce bottles of soda for $2.00. The number of bottles X bought from the machine on any day is a random variable with mean 40 and standard deviation 10. Let the random variable Y equal the total revenue from this machine on a given day. Assume that the machine works properly and that no sodas are stolen from the machine. What are the mean and standard deviation of Y?
(a) μY=$80.00 σY=$20.00
(b) μY=$80.00 σY=$200.00
(c) μY=$40.00 σY=$200.00
(d) μY=$40.00 σY=$10.00
(e) μY=$80.00 σY=$40.00
A
A study was conducted to analyze how many hours students spend studying for an exam. The random variable X represents the number of hours a randomly selected student studies. The probability distribution for X is given in the table below:
Hours(X) 0 1 2 3 4 5
P(X = x) 0.1 0.2 0.3 0.2 0.1 0.1
Part A
Verify that the given probability distribution is valid.
Part B
Calculate P(X ≤ 2), the probability that a randomly selected student studies at most 2 hours.
Part A
To verify that the probability distribution is valid, we need to check that:
Checking the first condition: All probabilities (0.1, 0.2, 0.3, 0.2, 0.1, 0.1) are between 0 and 1.
Checking the second condition: Sum of probabilities = 0.1 + 0.2 + 0.3 + 0.2 + 0.1 + 0.1 = 1.0
Since both conditions are satisfied, the given probability distribution is valid.
Part B
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.2 + 0.3 = 0.6
The probability that a randomly selected student studies at most 2 hours is 0.6 or 60%.
The total sales on a randomly-selected day at Joy's Toy Shop can be represented by the continuous random variable S, which has a Normal distribution wit a mean of $3600 and a standard deviation of $500. Find and interpret P(S>$4000)
Make a Normal graph with 2600, 3100, 3600, 4100, 4600.
Shade the area from 4000+
P(S>4000)= 0.2119
"The probability of total sales on a randomly selected day to be greater than $4000 is 0.2119"
Coffee Lover Mia’s Morning Routine
Mia is a coffee enthusiast who always brings a 1.2-pound bag of coffee on her weekend camping trips. The amount of coffee she uses each morning for brewing varies according to a Normal distribution with a mean of 0.18 pounds and a standard deviation of 0.04 pounds per day.
If Mia stays on a camping trip for 7 days, what is the probability that she will run out of coffee before the trip ends?
The probability that Mia will run out of coffee before the trip ends is approximately 71.46%.
A basketball player makes 75% of their free throws. If they take 5 shots, what is the probability they make exactly 3 of them?
(5
3) (0.75)3 (0.25)2
Suppose a student is randomly selected from a high school. Which of the following pairs of random variables are most likely independent?
(a) X = Student’s age; Y = Student’s height
(b) X = Student’s SAT Math score; Y = Student’s SAT Reading score
(c) X = Number of extracurricular activities a student participates in; Y = Student’s GPA
(d) X= Number of books a student reads per month; Y= Student’s writing proficiency score
(e) X = Student’s favorite color; Y = Student’s height
(e) X = Student’s favorite color; Y = Student’s height
A casino offers a game where a player pays $10 to roll a fair six-sided die. The player wins according to the following payout structure:
Calculate the expected value of X. Interpret this value in the context of the problem.
Expected value of X: E(X) = (-$5)(1/6) + (-$2)(1/3) + ($2)(1/3) + ($10)(1/6) = -$5/6 - $2/3 + $2/3 + $10/6 = -$5/6 + $10/6 = $5/6 ≈ $0.83
Interpretation: The player can expect to gain about $0.83 on average per play of the game. This means the game is slightly favorable to the player, and unfavorable to the casino in the long run.
A discrete random variable X has the following probability distribution:
X 0 1 2
P(X) .4 .3 .3
What is the variance of X?
Answer: A) 0.49
The manager of a children’s puppet theatre has determined that the h number of adult tickets he sells for a Saturday afternoon show is a random variable with a mean of 28.3 tickets and a standard deviation of 5.3 tickets. The mean number of children’s tickets he sells is 42.5, with a standard deviation of 8.1.
(a) The adult tickets sell for $10. Let A = the money he collects from adult tickets on a random Saturday. What are the mean and standard deviation of A?
(b) The children’s tickets sell for $6. Let T = the money he collects from all ticket sales. Assume that the number of tickets sold to adults is independent of the number sold to children. What are the mean and standard deviation of T?
(c) It costs $300 for the manager to put on each puppet show. Let P = the profit from a random show. What are the mean and standard deviation of P?
(a) Amew=$283 & Asd=$53
(b)Tmew=$538 & Tsd=$71.9094
(c)Pmew=$238 & Psd=$71.9094
According to Benford's law, the probability that the first digit of the amount of on a randomly chosen invoice is a 1 or a 2 is 0.477. Suppose you examine an SRS of 90 invoices from a vendor and find 29 that have first digits 1 or 2. Do you suspect that the invoice amounts are not genuine? Compute an appropriate probability to support your answer.
Because the probability of getting 29 or fewer invoices that begin with 1 or 2 is small, we can be suspicious that the invoice amounts are not genuine.
A researcher is testing a participant for extrasensory perception (ESP) by asking them to guess which one of six possible symbols—circle, star, triangle, diamond, heart, or square—is displayed on a hidden computer screen. The computer randomly selects a symbol each time, and all symbols are equally likely to be chosen. Suppose the participant does not have ESP and is guessing randomly.
What is the probability that the participant guesses incorrectly on the first three trials but correctly on the fourth?
A) 0.0579
B) 0.0347
C) 0.1157
D) 0.0682
B) 0.0965 ✅
An insurance company offers flood insurance policies to homeowners in a certain region. Based on historical data, the probability that a policyholder will file a claim during a one-year period is 0.08. The company charges an annual premium of $800 for each policy.
When a claim is filed, the insurance company pays out an amount that can be modeled by a normal distribution with a mean of $15,000 and a standard deviation of $3,500.
Part A
Let X be the random variable representing the company's net profit on a single policy in a year. Write an expression for X in terms of relevant variables, and identify the possible values of X.
Part B
Calculate the expected value of X for a single policy. Interpret this value in context.
Part A
Let C be the indicator random variable for whether a claim is filed (C = 1 if a claim is filed, C = 0 if not). Let A be the amount paid if a claim is filed.
X = $800 - C·A
Possible values of X:
Part B
E(X) = E($800 - C·A) = $800 - E(C·A) = $800 - E(C)·E(A) (since C and A are independent) = $800 - (0.08)·($15,000) = $800 - $1,200 = -$400
Interpretation: The insurance company expects to lose an average of $400 per policy. This means the premium of $800 is not high enough to cover the expected claim payouts.
A discrete random variable X represents the number of heads in 4 tosses of a fair coin. The probability distribution of X is as follows:
X 0 1 2 3 4
P(X) 1/16 4/16 6/16 4/16 1/16
a) Calculate the expected value, E(X), of the random variable X.
b) Calculate the variance, Var(X), of the random variable X.
a) E(X) = (0 * 1/16) + (1 * 4/16) + (2 * 6/16) + (3 * 4/16) + (4 * 1/16)
E(X) = 0 + 4/16 + 12/16 + 12/16 + 4/16 = 32/16 = 2
b) First, calculate E(X²):
E(X²) = (0² * 1/16) + (1² * 4/16) + (2² * 6/16) + (3² * 4/16) + (4² * 1/16)
E(X²) = 0 + 4/16 + 24/16 + 36/16 + 16/16 = 80/16 = 5
Then, calculate variance:
Var(X) = E(X²) − [E(X)]² = 5 − (2)² = 5 − 4 = 1
During the winter months, the temperatures in a downtown office building can drop significantly at night when the heating system is set to energy-saving mode. To ensure employees arrive to a comfortable environment, the facility manager sets the thermostat to maintain a temperature of 68°F. Additionally, a digital thermostat records the office temperature at midnight, but it records the temperature in degrees Celsius.
Based on several years of data, the temperature T in the office at midnight on a randomly selected night follows a Normal distribution with a mean of 19°C and a standard deviation of 1.8°C.
(a) What is the probability that on a randomly chosen night, the recorded office temperature is below 17°C?
(b) What is the probability that the office temperature is above 21°C?
(a) 13.35% chance that the office temperature is below 17°C.
(b) 13.35% chance that the office temperature is above 21°C.
A company has 50 employees, 10 of whom have access to highly sensitive information. As part of a routine drug testing program, the company randomly selects 10 employees for testing. Employees were surprised when none of the 10 selected employees had access to highly sensitive information.
Can we use a binomial distribution to approximate this probability? Justify your answer.
Can we use a binomial distribution to approximate this probability? Justify your answer.
A binomial distribution is not appropriate at all in this case because the selections are highly dependent and the population size is too small relative to the sample
In the Japanese game show "Sushi Roulette," the contestant spins a large wheel divided into 15 equal sections. Ten of the sections have a sushi roll, and five have a "wasabi bomb." When the wheel stops, the contestant must eat whatever food is on that section. To win the game, the contestant must eat one wasabi bomb. Find the probability that it takes 3 or fewer spins for the contestant to get a wasabi bomb.
(a) 0.481
(b) 0.556
(c) 0.704
(d) 0.813
(e) 0.925
A manufacturing plant produces electronic components. Each component has a 5% probability of being defective, independent of other components. The plant ships the components in boxes of 25.
Part A
Let X be the random variable representing the number of defective components in a box. Identify the probability distribution of X and its parameters.
Part B
Calculate the probability that a box contains exactly 2 defective components.
Part C
Calculate the probability that a box contains at least 3 defective components.
Part D
The plant manager wants to establish a quality control policy. Under this policy, a box will be inspected if the probability of having more than 1 defective component exceeds 0.3. Based on your calculations, should boxes be inspected under this policy? Justify your answer.
Part E
The quality control team is considering improving the manufacturing process to reduce the probability of a defective component to 2%. If this improvement is implemented, what would be the expected number of defective components per box? What would be the standard deviation?
Part A
X follows a binomial distribution with n = 25 and p = 0.05. X ~ Bin(25, 0.05)
Part B
P(X = 2) = C(25,2) · (0.05)² · (0.95)²³ = (25!/(2!·23!)) · (0.05)² · (0.95)²³ = 300 · 0.0025 · 0.3104 ≈ 0.2329
The probability of exactly 2 defective components is approximately 0.2329 or 23.29%.
Part C
P(X ≥ 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 1 - [(0.95)²⁵ + 25·(0.05)·(0.95)²⁴ + 300·(0.05)²·(0.95)²³] = 1 - [0.2774 + 0.3650 + 0.2329] = 1 - 0.8753 ≈ 0.1247
The probability of at least 3 defective components is approximately 0.1247 or 12.47%.
Part D
We have already calculated that: P(X > 1) = 1 - [P(X = 0) + P(X = 1)] = 1 - [0.2774 + 0.3650] = 1 - 0.6424 ≈ 0.3576
Since 0.3576 > 0.3, the probability of having more than 1 defective component exceeds the threshold of 0.3. Therefore, boxes should be inspected under this policy.
Part E
If p = 0.02, then X ~ Bin(25, 0.02)
Expected value: E(X) = n·p = 25·0.02 = 0.5 Standard deviation: SD(X) = √(n·p·(1-p)) = √(25·0.02·0.98) ≈ 0.7
With the improved process, we would expect an average of 0.5 defective components per box, with a standard deviation of approximately 0.7 components.