Statistics MCQs
Topic Notes: Statistics
MCQs and preparation resources for competitive exams, covering important concepts, past papers, and detailed explanations.
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
1
How is the variance of a scaled random variable, Var(kY), mathematically expressed?
Answer:
k2 Var(Y)
According to the properties of variance, when a random variable Y is multiplied by a constant k, the variance of the new variable kY is equal to k squared multiplied by the variance of the original variable Y. This demonstrates that scaling a variable by k increases its variance by a factor of k squared.
2
What is the variance of the linear transformation 2X + 3, expressed in terms of the variance of X?
Answer:
4 Var(X)
According to the properties of variance, Var(aX + b) = a^2 * Var(X). Here, a = 2 and b = 3. Therefore, Var(2X + 3) = 2^2 * Var(X) = 4 * Var(X). The constant term b does not affect the variance because shifting the data does not change the spread.
3
What is the result of multiplying the number of units by the profit per unit and the probability of that outcome?
Answer:
expected profit
Expected profit is calculated by taking the sum of all possible profit outcomes, each multiplied by its respective probability of occurrence. This represents the long-term average profit one would expect if the scenario were repeated many times.
4
What is the formula for the expected mean of a hypergeometric distribution?
Answer:
Nk/N
The mean of a hypergeometric distribution, where N is the population size, K is the number of successes in the population, and n is the sample size, is given by n multiplied by (K/N). This represents the expected number of successes in the sample drawn without replacement.
5
For a discrete random variable X, how is the expected value E(X) calculated?
Answer:
∑x.P(x)
The expected value, or mean, of a discrete random variable X is defined as the weighted average of all possible values that X can take, where the weights are the probabilities of those values. Mathematically, this is expressed as the sum of the product of each value x and its corresponding probability P(x), denoted as E(X) = ∑ x * P(x).
6
What is the expected value of the deviation of a random variable X from its mean?
Answer:
0
The expected value of the deviation of a random variable from its mean is defined as E[X - E(X)]. By the linearity of expectation, this becomes E(X) - E(E(X)). Since E(X) is a constant, E(E(X)) = E(X), resulting in E(X) - E(X) = 0. This property reflects that the mean is the center of gravity of the distribution.
7
What is the variance of the linear transformation (a + bX), where 'a' and 'b' are constants and 'X' is a random variable?
Answer:
b2Var(X)
The variance of a constant 'a' is zero. For a linear transformation (a + bX), the variance is calculated as Var(a + bX) = Var(a) + Var(bX). Since Var(a) = 0 and Var(bX) = b^2 * Var(X), the resulting variance is b^2 * Var(X). This property demonstrates that variance is invariant to shifts in origin but sensitive to changes in scale.
8
Given a random variable X, how is the variance Var(X) defined in terms of the expectation of squared deviations?
Answer:
(x – 4)2
The variance is defined as E[(X - E[X])^2]. The provided answer D appears to be a typographical representation of E[(X - μ)^2], where μ is the mean. We preserve the answer key while noting the likely notation error in the source material.
9
Given daily product demands of 21, 19, and 22 units with associated probabilities of 0.29, 0.40, and 0.35 respectively, and a profit of $0.50 per unit, what are the expected profits for each scenario?
Answer:
3.045, 3.8, 3.85
To calculate the expected profit for each outcome, multiply the demand by the profit per unit ($0.50) and then by the probability of that demand occurring. For 21 units: 21 * 0.50 * 0.29 = 3.045. For 19 units: 19 * 0.50 * 0.40 = 3.8. For 22 units: 22 * 0.50 * 0.35 = 3.85.
10
What term describes the arithmetic mean of all potential outcomes in a probability distribution?
Answer:
expected value
The expected value is the long-run average or mean value of a random variable over many repeated trials. It is calculated by summing the products of each possible outcome and its associated probability, representing the theoretical center of a probability distribution.