Discrete Random Variables
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
Why do we use cdf and never pdf when working with continuous random variables?
that event. Specific points have no area under a curve, which is what pdf is used for.
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
Which of the following are continuous random variables (select all)
A) Shoe Size
B)100m dash time
C) Wingspan of professional basketball players
D) Number of household pets
E) Time spent on bathroom pass
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
What conditions must be met for a distribution to be geometric?
BITS
Binary
Independent
Trials
Success
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 4 of them?
binompdf (5, .75, 4)
=.396
5% of the tomatoes at a farmer’s market have imperfections on them. You randomly choose one tomato at a time until you find one with an imperfection. What is the probability you find your first bad tomato within five tomatoes?
P(x<=5)
geometcdf(5, .05)
=.226
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
The probability that a particular student passes an assessment is 70%. What is the probability that the student
takes exactly three attempts to pass the assessment?
.063
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.
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.
0.704
The local department store is holding a promotion: with every purchase, you get a free mystery goodie bag. There are various prizes in the bag but you are interested in the $20 gift certificate in them. The $20 gift certificate is only in 10% of goodie bags created. Let X = the number of goodies bags that must be open until the gift certificate is found.
Determine what type of distribution this is, then find the expeted value.