Q1: A discrete random variable is one that: (a) Can take any value within an interval (b) Can take only specific, countable values (c) Must always be positive (d) Has an infinite number of possible values
Solution:
Ans: (b) Explanation: A discrete random variable can take only specific, countable values (like 0, 1, 2, 3, etc.). Option (a) describes a continuous random variable. Options (c) and (d) are incorrect because discrete variables can be negative and may have finite values.
Q2: The sum of all probabilities in a probability distribution must equal: (a) 0 (b) 0.5 (c) 1 (d) The number of outcomes
Solution:
Ans: (c) Explanation: One of the fundamental properties of a probability distribution is that the sum of all probabilities must equal 1 (or 100%). This ensures that one of the possible outcomes must occur.
Q3: If X is a discrete random variable with values {1, 2, 3} and P(X = 1) = 0.3 and P(X = 2) = 0.5, then P(X = 3) equals: (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.8
Solution:
Ans: (b) Explanation: Since the sum of all probabilities must equal 1, we have: \(P(X = 1) + P(X = 2) + P(X = 3) = 1\) \(0.3 + 0.5 + P(X = 3) = 1\) \(P(X = 3) = 1 - 0.8 = 0.2\)
Q4: The expected value E(X) of a discrete random variable represents: (a) The most frequently occurring value (b) The weighted average of all possible values (c) The middle value when data is ordered (d) The difference between maximum and minimum values
Solution:
Ans: (b) Explanation: The expected value \(E(X)\) is the weighted average of all possible values, where each value is weighted by its probability. Option (a) describes the mode, option (c) describes the median, and option (d) describes the range.
Q5: Given the probability distribution: P(X = 0) = 0.4, P(X = 1) = 0.35, P(X = 2) = 0.25, what is E(X)? (a) 0.75 (b) 0.85 (c) 1.00 (d) 1.15
Solution:
Ans: (b) Explanation: The expected value is calculated as: \(E(X) = \sum x \cdot P(X = x)\) \(E(X) = 0(0.4) + 1(0.35) + 2(0.25)\) \(E(X) = 0 + 0.35 + 0.50 = 0.85\)
Q6: The variance of a discrete random variable measures: (a) The average value of the variable (b) The spread or dispersion of values around the mean (c) The probability of the most likely outcome (d) The total of all possible values
Solution:
Ans: (b) Explanation:Variance measures how spread out the values of a random variable are from the expected value (mean). A larger variance indicates greater dispersion, while a smaller variance indicates values are closer to the mean.
Q7: If E(X) = 5 and Var(X) = 9, what is the standard deviation of X? (a) 2 (b) 3 (c) 4 (d) 14
Solution:
Ans: (b) Explanation: The standard deviation is the square root of the variance: \(\sigma = \sqrt{Var(X)} = \sqrt{9} = 3\)
Q8: Which of the following is NOT a valid probability distribution for a discrete random variable? (a) P(X = 1) = 0.3, P(X = 2) = 0.4, P(X = 3) = 0.3 (b) P(X = 0) = 0.5, P(X = 1) = 0.3, P(X = 2) = 0.2 (c) P(X = 1) = 0.6, P(X = 2) = 0.5, P(X = 3) = 0.1 (d) P(X = 0) = 0.25, P(X = 1) = 0.25, P(X = 2) = 0.5
Solution:
Ans: (c) Explanation: For a valid probability distribution, the sum of all probabilities must equal 1, and each probability must be between 0 and 1. In option (c), the sum is \(0.6 + 0.5 + 0.1 = 1.2\), which exceeds 1, making it invalid.
## Section B: Fill in the Blanks
Q9:A random variable that can only take on a countable number of distinct values is called a __________ random variable.
Solution:
Ans: discrete Explanation: A discrete random variable has countable, separate values (such as the number of students in a class or the result of rolling a die), as opposed to continuous random variables which can take any value in an interval.
Q10:The formula for the expected value of a discrete random variable X is E(X) = __________.
Solution:
Ans: \(\sum x \cdot P(X = x)\) or \(\sum xP(x)\) Explanation: The expected value is calculated by multiplying each possible value by its probability and summing all these products. This gives the long-run average value of the random variable.
Q11:For any probability distribution, each individual probability must be between __________ and __________ inclusive.
Solution:
Ans: 0 and 1 Explanation: Every probability value must satisfy \(0 \leq P(X = x) \leq 1\). Probabilities cannot be negative or greater than 1, as they represent the likelihood of an event occurring.
Q12:The variance of a discrete random variable is calculated using the formula Var(X) = __________.
Solution:
Ans: \(\sum (x - \mu)^2 \cdot P(X = x)\) or \(E(X^2) - [E(X)]^2\) Explanation: The variance measures spread and can be calculated as the expected value of squared deviations from the mean, or equivalently as \(E(X^2) - [E(X)]^2\).
Q13:The standard deviation is the __________ of the variance.
Solution:
Ans: square root Explanation: The standard deviation \(\sigma = \sqrt{Var(X)}\) provides a measure of spread in the same units as the original data, making it easier to interpret than variance.
Q14:If a discrete random variable X has only two possible outcomes, it is called a __________ random variable.
Solution:
Ans: Bernoulli or binomial Explanation: A Bernoulli random variable has exactly two outcomes (success/failure), and is the simplest case of a binomial distribution with a single trial.
## Section C: Word Problems
Q15:A basketball player has a probability of 0.6 of making a free throw. Let X represent the number of successful free throws out of 1 attempt. Construct the probability distribution table for X and find the expected value.
Solution:
Ans: Probability Distribution Table: \(X = 0\): \(P(X = 0) = 0.4\) \(X = 1\): \(P(X = 1) = 0.6\)
Final Answer: The expected value is 0.6 successful free throws.
Q16:A game involves rolling a fair six-sided die. You win $4 if you roll a 6, $2 if you roll a 4 or 5, and $0 for any other outcome. Let X represent your winnings. Find the expected value of your winnings.
Final Answer: The expected winnings are $1.33 (or $4/3).
Q17:A discrete random variable Y has the following probability distribution: P(Y = 2) = 0.2, P(Y = 4) = 0.3, P(Y = 6) = 0.5. Calculate the variance of Y.
Q18:A quality control inspector examines items from a production line. The number of defective items X in a sample of 5 items has the following distribution: P(X = 0) = 0.50, P(X = 1) = 0.30, P(X = 2) = 0.15, P(X = 3) = 0.05. Find the expected number of defective items and the standard deviation.
Standard Deviation: \(\sigma = \sqrt{0.7875} \approx 0.887\)
Final Answer: Expected number of defective items is 0.75, and the standard deviation is approximately 0.89 items.
Q19:A student takes a multiple-choice quiz with 3 questions. Each question has 4 options, and the student guesses randomly on each question. Let X represent the number of correct answers. Create the probability distribution for X and calculate P(X ≥ 2).
Solution:
Ans: This is a binomial situation with \(n = 3\) trials and probability of success \(p = \frac{1}{4} = 0.25\).
Using binomial probability formula \(P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\):
Final Answer: The probability of getting at least 2 correct answers is 0.157 or approximately 15.7%.
Q20:A vending machine dispenses snacks, and the number of snacks purchased per hour (X) follows this distribution: P(X = 5) = 0.15, P(X = 10) = 0.35, P(X = 15) = 0.40, P(X = 20) = 0.10. If the profit per snack is $0.50, what is the expected hourly profit from the vending machine?
Solution:
Ans: First, find the expected number of snacks sold per hour: \(E(X) = 5(0.15) + 10(0.35) + 15(0.40) + 20(0.10)\) \(E(X) = 0.75 + 3.5 + 6.0 + 2.0 = 12.25\) snacks
Expected hourly profit: Expected profit = Expected number of snacks × Profit per snack Expected profit = \(12.25 \times 0.50 = 6.125\)
Final Answer: The expected hourly profit is $6.13 (or $6.125).
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