Mathematics

Probability Distributions

488 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

Multiple choice

A random variable X has a probability mass function given by P(X = x) = 1/6 for x = 1, 2, 3, 4, 5, 6. Find the expected value of X.

  1. 3.5

  2. 4

  3. 4.5

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expected value of X is given by E(X) = Σ[x * P(X = x)] for all possible values of X. So, E(X) = 1 * (1/6) + 2 * (1/6) + 3 * (1/6) + 4 * (1/6) + 5 * (1/6) + 6 * (1/6) = 3.5.

Multiple choice

A company claims that their new product will last for at least 100 hours. A consumer group tests the product and finds that the average lifetime is 95 hours with a standard deviation of 10 hours. Assuming the lifetime follows a normal distribution, what is the probability that a randomly selected product will last for less than 80 hours?

  1. 0.0228

  2. 0.1587

  3. 0.3413

  4. 0.8413
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can use the standard normal distribution (Z) to calculate this probability: P(X < 80) = P((X - 95)/10 < (80 - 95)/10) = P(Z < -1.5) = 0.0228.

Multiple choice

What is the probability of selecting a specific element from a population of size N using simple random sampling?

  1. 1/N

  2. N/1

  3. 1/(N-1)

  4. (N-1)/N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In simple random sampling, each element in the population has an equal chance of being selected, so the probability of selecting a specific element is 1/N.

Multiple choice

What is the probability of exactly k events occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda?

  1. P(X = k) = (lambda^k * e^(-lambda)) / k!

  2. P(X = k) = (lambda^k * e^(-lambda)) / (k + 1)!

  3. P(X = k) = (lambda^k * e^(-lambda)) / k

  4. P(X = k) = (lambda^k * e^(-lambda)) / (k - 1)!
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of exactly k events occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda, is given by the Poisson distribution formula: P(X = k) = (lambda^k * e^(-lambda)) / k!

Multiple choice

What is the probability of no events occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda?

  1. P(X = 0) = e^(-lambda)

  2. P(X = 0) = 1 - e^(-lambda)

  3. P(X = 0) = lambda * e^(-lambda)

  4. P(X = 0) = 1 - lambda * e^(-lambda)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of no events occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda, is given by P(X = 0) = e^(-lambda).

Multiple choice

What is the probability of at least one event occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda?

  1. P(X >= 1) = 1 - e^(-lambda)

  2. P(X >= 1) = e^(-lambda)

  3. P(X >= 1) = lambda * e^(-lambda)

  4. P(X >= 1) = 1 - lambda * e^(-lambda)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of at least one event occurring in a fixed interval of time or space, given that the average rate of occurrence is lambda, is given by P(X >= 1) = 1 - e^(-lambda).

Multiple choice

A call center receives an average of 10 calls per hour. What is the probability that the call center receives exactly 12 calls in the next hour?

  1. 0.125

  2. 0.25

  3. 0.375

  4. 0.5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let X be the number of calls received in the next hour. Then X follows a Poisson distribution with a mean of 10. The probability of receiving exactly 12 calls is given by P(X = 12) = (10^12 * e^(-10)) / 12! = 0.125.

Multiple choice

A machine produces defective items with a probability of 0.05. What is the probability that the machine produces exactly 2 defective items in the next 100 items?

  1. 0.102

  2. 0.204

  3. 0.306

  4. 0.408
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let X be the number of defective items produced in the next 100 items. Then X follows a Poisson distribution with a mean of 5 (0.05 * 100). The probability of producing exactly 2 defective items is given by P(X = 2) = (5^2 * e^(-5)) / 2! = 0.204.

Multiple choice

A store sells an average of 20 items per day. What is the probability that the store sells exactly 25 items on a particular day?

  1. 0.135

  2. 0.271

  3. 0.406

  4. 0.542
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let X be the number of items sold on a particular day. Then X follows a Poisson distribution with a mean of 20. The probability of selling exactly 25 items is given by P(X = 25) = (20^25 * e^(-20)) / 25! = 0.271.

Multiple choice

A company receives an average of 50 emails per hour. What is the probability that the company receives no emails in the next 15 minutes?

  1. 0.067

  2. 0.135

  3. 0.203

  4. 0.271
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let X be the number of emails received in the next 15 minutes. Then X follows a Poisson distribution with a mean of 12.5 (50 * 15 / 60). The probability of receiving no emails is given by P(X = 0) = e^(-12.5) = 0.203.

Multiple choice

A doctor sees an average of 10 patients per day. What is the probability that the doctor sees at least one patient on a particular day?

  1. 0.995

  2. 0.950

  3. 0.900

  4. 0.850
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let X be the number of patients seen on a particular day. Then X follows a Poisson distribution with a mean of 10. The probability of seeing at least one patient is given by P(X >= 1) = 1 - P(X = 0) = 1 - e^(-10) = 0.995.

Multiple choice

A machine produces bolts with a probability of 0.99 of being defective. What is the probability that the machine produces at most 2 defective bolts in the next 100 bolts?

  1. 0.904

  2. 0.950

  3. 0.990

  4. 0.999
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let X be the number of defective bolts produced in the next 100 bolts. Then X follows a Poisson distribution with a mean of 99 (0.99 * 100). The probability of producing at most 2 defective bolts is given by P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2) = e^(-99) + 99 * e^(-99) + (99^2 * e^(-99)) / 2! = 0.990.

Multiple choice

A company receives an average of 20 phone calls per hour. What is the probability that the company receives between 15 and 25 phone calls in the next hour?

  1. 0.242

  2. 0.368

  3. 0.494

  4. 0.620
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let X be the number of phone calls received in the next hour. Then X follows a Poisson distribution with a mean of 20. The probability of receiving between 15 and 25 phone calls is given by P(15 <= X <= 25) = P(X = 15) + P(X = 16) + ... + P(X = 25) = 0.494.

Multiple choice

A store sells an average of 50 items per day. What is the probability that the store sells more than 60 items on a particular day?

  1. 0.067

  2. 0.135

  3. 0.203

  4. 0.271
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let X be the number of items sold on a particular day. Then X follows a Poisson distribution with a mean of 50. The probability of selling more than 60 items is given by P(X > 60) = 1 - P(X <= 60) = 1 - (P(X = 0) + P(X = 1) + ... + P(X = 60)) = 0.203.

Multiple choice

A doctor sees an average of 15 patients per day. What is the probability that the doctor sees exactly 20 patients on a particular day?

  1. 0.102

  2. 0.204

  3. 0.306

  4. 0.408
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let X be the number of patients seen on a particular day. Then X follows a Poisson distribution with a mean of 15. The probability of seeing exactly 20 patients is given by P(X = 20) = (15^20 * e^(-15)) / 20! = 0.204.