Statistics: Probability Distributions
Modeling outcomes with discrete and continuous distributions
Statistics: Probability Distributions
Modeling outcomes with discrete and continuous distributions
Statistics - Grade 9-12
- 1
A random variable X has the following probability distribution: P(X = 0) = 0.10, P(X = 1) = 0.25, P(X = 2) = 0.40, and P(X = 3) = 0.25. Verify that this is a valid probability distribution.
A discrete probability distribution must have probabilities that add to exactly 1.
This is a valid probability distribution because all probabilities are between 0 and 1, and the sum is 0.10 + 0.25 + 0.40 + 0.25 = 1.00. - 2
Using the distribution P(X = 0) = 0.10, P(X = 1) = 0.25, P(X = 2) = 0.40, and P(X = 3) = 0.25, find the expected value of X.
Multiply each value of X by its probability, then add the products.
The expected value is E(X) = 0(0.10) + 1(0.25) + 2(0.40) + 3(0.25) = 0 + 0.25 + 0.80 + 0.75 = 1.80. The expected value of X is 1.80. - 3
A fair six-sided die is rolled once. Let X be the number rolled. What type of probability distribution does X have, and what is P(X = 4)?
X has a discrete uniform distribution because each outcome 1 through 6 is equally likely. The probability P(X = 4) is 1/6, or about 0.167. - 4
A bag contains 5 red marbles and 3 blue marbles. One marble is selected at random. Let X = 1 if the marble is red and X = 0 if the marble is blue. Give the probability distribution of X.
Match each value of X to the event it represents.
There are 8 marbles total. Since P(red) = 5/8 and P(blue) = 3/8, the distribution is P(X = 1) = 5/8 and P(X = 0) = 3/8. - 5
A basketball player makes 70% of free throws. If the player shoots 5 free throws, what is the probability that the player makes exactly 4? Assume each shot is independent.
Use the binomial formula P(X = k) = C(n,k)p^k(1 - p)^(n - k).
This is a binomial situation with n = 5, p = 0.70, and k = 4. The probability is C(5,4)(0.70)^4(0.30)^1 = 5(0.2401)(0.30) = 0.360. The probability is about 0.360. - 6
A multiple-choice quiz has 10 questions, each with 4 answer choices. A student guesses on every question. Let X be the number of correct answers. Identify the distribution and state n and p.
X follows a binomial distribution because there is a fixed number of independent trials, each trial has two outcomes, and the probability of success is constant. The parameters are n = 10 and p = 1/4, or 0.25. - 7
For a binomial random variable with n = 8 and p = 0.25, find the mean and standard deviation.
For a binomial distribution, the mean is np and the standard deviation is sqrt(np(1 - p)).
The mean is np = 8(0.25) = 2. The standard deviation is sqrt(np(1 - p)) = sqrt(8(0.25)(0.75)) = sqrt(1.5), which is about 1.225. - 8
A factory finds that 2% of light bulbs are defective. If 12 bulbs are randomly selected, what is the probability that none are defective? Assume independence.
If none are defective, all 12 bulbs are not defective.
This is a binomial probability with n = 12, p = 0.02, and k = 0. The probability is C(12,0)(0.02)^0(0.98)^12 = (0.98)^12, which is about 0.785. - 9
A spinner has four equal sections labeled A, B, C, and D. It is spun 20 times. Which distribution would model the number of times the spinner lands on A, and why?
The number of times the spinner lands on A can be modeled by a binomial distribution because there are 20 fixed spins, each spin is independent, each spin can be counted as A or not A, and P(A) = 1/4 stays the same. - 10
In a geometric distribution, the probability of success on each trial is p = 0.20. What is the probability that the first success occurs on the 4th trial?
The first three trials must be failures, and the fourth trial must be a success.
For a geometric distribution, P(X = 4) = (1 - p)^3p = (0.80)^3(0.20) = 0.1024. The probability is about 0.102. - 11
A customer service center receives an average of 6 calls per hour. If calls follow a Poisson distribution, what is the probability of receiving exactly 4 calls in one hour?
Use P(X = k) = e^(-lambda)lambda^k/k! for a Poisson distribution.
Using the Poisson formula with lambda = 6 and k = 4, P(X = 4) = e^(-6)6^4/4!. This is about 0.134. The probability of exactly 4 calls is about 0.134. - 12
The number of emails a teacher receives during lunch follows a Poisson distribution with a mean of 3. What is the variance of this distribution?
For a Poisson distribution, the variance equals the mean. Since the mean is 3, the variance is 3. - 13
Scores on a test are normally distributed with a mean of 75 and a standard deviation of 10. Find the z-score for a student who scored 90.
Use z = (x - mean)/standard deviation.
The z-score is z = (90 - 75)/10 = 15/10 = 1.5. The student scored 1.5 standard deviations above the mean. - 14
Heights of adult women in a town are normally distributed with a mean of 64 inches and a standard deviation of 3 inches. About what percent of adult women are between 61 and 67 inches tall?
Use the empirical rule for a normal distribution.
The values 61 and 67 are one standard deviation below and above the mean because 64 - 3 = 61 and 64 + 3 = 67. By the empirical rule, about 68% of adult women are between 61 and 67 inches tall. - 15
A continuous random variable is uniformly distributed from 0 to 10. What is the probability that a randomly selected value is between 3 and 7?
For a continuous uniform distribution, compare the length of the desired interval to the full interval.
For a uniform distribution, probability equals interval length divided by total length. The interval from 3 to 7 has length 4, and the total interval from 0 to 10 has length 10. The probability is 4/10 = 0.400.