Mean and Variance of Random variables MCQ Quiz - Objective Question with Answer for Mean and Variance of Random variables - Download Free PDF
Last updated on May 15, 2025
Latest Mean and Variance of Random variables MCQ Objective Questions
Mean and Variance of Random variables Question 1:
A random variable X has the following probability distribution:
X | 0 | 1 | 2 | otherwise |
P(X) | k | 2k | 3k | 0 |
Then:
(A) \(\mathrm{k}=\frac{1}{6}\)
(B) \(P(X<2)=\frac{1}{2}\)
(C) \(\mathrm{E}(\mathrm{X})=\frac{3}{4}\)
(D) \(P(1
Choose the correct answer from the options given below:
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 1 Detailed Solution
Concept:
Probability Distribution:
- A probability distribution describes how the probabilities are distributed over the values of the random variable.
- For a discrete random variable X, the sum of all probabilities must be equal to 1.
- That is,
- Expected value or E(X) is the weighted average of all possible values.
- Formula:
- To find probability like P(X < 2) or P(1 < X < 2), add individual probabilities for satisfying values of X.
Calculation:
Given,
P(X=0) = k, P(X=1) = 2k, P(X=2) = 3k
⇒ Total Probability = k + 2k + 3k = 6k
⇒ 6k = 1
⇒ k = 1 / 6
⇒ P(X < 2) = P(X = 0) + P(X = 1)
⇒ P(X < 2) = k + 2k = 3k = 3 × (1/6) = 1/2
⇒ E(X) = 0 × k + 1 × 2k + 2 × 3k
⇒ E(X) = 0 + 2k + 6k = 8k = 8 × (1/6) = 4/3
⇒ P(1 < X < 2) = P(X = 2) = 3k = 3 × (1/6) = 1/2
∴ Correct statements are (A) and (B) only.
Mean and Variance of Random variables Question 2:
Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is :
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 2 Detailed Solution
Calculation
Probability distribution
\(\begin{array}{c|c} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \\ \hline \mathrm{x}=0 & \frac{7 \mathrm{C}_{2}}{10 \mathrm{C}_{2}}=\frac{42}{90} \\ \hline \mathrm{x}=1 & \frac{7 \mathrm{C}_{1} \times 3 \mathrm{C}_{1}}{10 \mathrm{C}_{2}}=\frac{42}{90} \\ \hline \mathrm{x}=2 & \frac{3 \mathrm{C}_{2}}{10 \mathrm{C}_{2}}=\frac{6}{90} \end{array}\)
Now,
\(\mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{42}{90}+\frac{12}{90}=\frac{54}{90}\)
\(\sigma^{2}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{1}^{2}-\mu^{2}=\frac{42}{90}+\frac{24}{90}-\left(\frac{54}{90}\right)^{2}\)
⇒ \(\frac{66}{90}-\left(\frac{54}{90}\right)^{2}\)
⇒ \(\sigma^{2} \Rightarrow \frac{28}{75} \)
Hence option 1 is correct
Mean and Variance of Random variables Question 3:
A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ + σ2) is :
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 3 Detailed Solution
Calculation
HHH → 0
HHT → 0
HTH → 1
HTT → 0
THH → 1
THT → 1
TTH → 1
TTT → 0
Probability distribution
\(\begin{array}{c|c|c} \mathrm{x}_{\mathrm{i}} & 0 & 1 \\ \hline \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right) & 1 / 2 & 1 / 2 \end{array}\)
\(\mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{2}\)
\(\sigma^{2}=\sum x_{i}^{2} p_{i}-\mu^{2}\)
= \(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(64\left(\mu+\sigma^{2}\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48\)
Hence option 2 is correct
Mean and Variance of Random variables Question 4:
If the probability distribution of X is :
X | 1 | 2 | 3 | 4 | 5 | 6 |
P(X) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) |
then Var.(X) = _________.
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 4 Detailed Solution
Calculation
E(X) = \(P(X)\cdot \frac{n}{2}\cdot (n+1) =\frac{1}{6}\cdot \frac{6}{2}\cdot 7 =\frac{7}{2}\)
E(X2) = \(\frac{1}{6}\times \frac{6}{6} \times 7\times 13\) = \(\frac{91}{6}\)
Var.(X) = {E(X)}2 - E(X2) = \(\frac{49}{4}-\frac{91}{6} = \frac{35}{12}\)
Hence option 3 is correct
Mean and Variance of Random variables Question 5:
For the probability distribution of a discrete random variable X as given below, the mean of X is
X = x | -2 | -1 | 0 | 1 | 2 | 3 |
P (X = x) | \(\frac{1}{10}\) | \(K+\frac{2}{10}\) | \(K+\frac{3}{10}\) | \(K+\frac{3}{10}\) | \(K+\frac{4}{10}\) | \(K+\frac{2}{10}\) |
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 5 Detailed Solution
Concept:
Mean (Expected Value) of a Discrete Random Variable:
- The mean or expected value of a discrete random variable X is the weighted average of all possible values that X can take.
- It is calculated as: E(X) = Σ [x × P(X = x)]
- All probabilities must sum up to 1: Σ P(X = x) = 1
- The result of E(X) represents the average outcome over many repetitions of the random experiment.
Calculation:
Given,
Probability distribution is:
X = x | -2 | -1 | 0 | 1 | 2 | 3 |
P (X = x) | \(\frac{1}{10}\) | \(K+\frac{2}{10}\) | \(K+\frac{3}{10}\) | \(K+\frac{3}{10}\) | \(K+\frac{4}{10}\) | \(K+\frac{2}{10}\) |
Step 1: Find the total probability
Sum of all probabilities = 1
K/10 + 2/10 + K/10 + 3/10 + (K + 2)/10 + K/10 = 1
⇒ (4K + 7)/10 = 1
⇒ 4K + 7 = 10
⇒ 4K = 3
⇒ K = 3/4
Step 2: Compute the expected value E(X)
E(X) = Σ [x × P(X = x)]
E(X) = (-2 × 3/40) + (-1 × 2/10) + (0 × 3/40) + (1 × 3/10) + (2 × 11/40) + (3 × 3/40)
⇒ E(X) = -6/40 - 8/40 + 0 + 12/40 + 22/40 + 9/40
⇒ E(X) = (-6 - 8 + 12 + 22 + 9)/40
⇒ E(X) = 29/40
⇒ E(X) = 15/15 - 14/15 = 4/5
∴ Mean of X is 4/5.
Top Mean and Variance of Random variables MCQ Objective Questions
Let Z denote the number of hours you study on a monday. Also it is known that
P(Z = z) = \(\rm \left\{\begin{matrix} 0.4 & if z = 0\\ kz & if z = 1 \: or \: 2\\ 0 & otherwise \end{matrix}\right. \)
where k is constant.
What is the probability that you study atleast two hours?
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 6 Detailed Solution
Download Solution PDFCalculation:
Probability Distribution of Z is
Z | 0 | 1 | 2 | otherwise |
P(Z) | 0.4 | k | 2k | 0 |
0.4 + k + 2k = 1
3k = 1 - 0.4
k = 0.2
P(you study atleast two hours)
P(Z ≥ 2) = P(2) + P(3) + P(4) + .....
= 2k
= 2 × 0.2
P(Z ≥ 2) = 0.4
A random variable X has the distribution law as given below:
X |
1 |
2 |
3 |
P(X = x) |
0.3 |
0.4 |
0.3 |
The variance of the distribution is:
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 7 Detailed Solution
Download Solution PDFConcept:
For a random variable X = xi with probabilities P(X = x) = pi:
- Mean/Expected Value: μ = ∑pixi.
- Variance: Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2.
- Standard Deviation: σ = \(\rm \sqrt{Var(X)}\).
Calculation:
We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3.
Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3)
= 0.3 + 0.8 + 0.9
= 2
And, ∑pi(xi)2 = (0.3 × 12) + (0.4 × 22) +(0.3 × 32)
= 0.3 + 1.6 + 2.7
= 4.6
∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6.
The variance of the distribution is Var(X) = 0.6.
A box contains 10 blue and 5 red balloons.If 5 balloons are randomly drawn, one-by-one, with replacement, then the variance of the number of blue balloons drawn is?
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 8 Detailed Solution
Download Solution PDFFormula used:
Variance = npq
Where,
p = probability of success,
q = the probability of failure,
n = the total no. of attempts
Calculation:
p (probability of drawing a blue balloon) = \(\rm \frac{10}{15} = \frac{2}{3} \)
Here, n = 5, p = \(\rm \frac{2}{3} \) ,q = \(\rm \frac{1}{3} \)
Then, variance = \(\rm 5 \times \frac{2}{3} \times \frac{1}{3} \) = \(\rm \frac{10}{9} \)
The variance of the number of blue balloons drawn is \(\rm \frac{10}{9} \)
The mean of the probability distribution function given by the following table will be:
x | 1 | 2 | 3 | 4 | 5 |
P(x) | 0.2 | 0.35 | 0.25 | 0.15 | 0.05 |
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 9 Detailed Solution
Download Solution PDFConcept:
Expected Value (or mean) of a Discrete Random Variable:
For a discrete random variable, the expected value usually denoted as μ or E(X), is calculated using
μ = E(X) = ∑ xiP(xi)
1. The formula means that we multiply each value x, in the support by its respective probability P(x) and then add them all together.
2. It can be seen as an average value but weighted by the likelihood of the value.
Calculation:
We know that, mean
μ = E(X) = ∑ xiP(xi)
⇒ E(X) = 1 × 0.2 + 2 × 0.35 + 3 × 0.25 + 4 × 0.15 + 5 × 0.05
⇒ E(X) = 0.2 + 0.7 + 0.75 + 0.6 + 0.25
⇒ E(X) = 2.5
Additional Information
The variance of a discrete random variable is given by:
Var(X) = σ2(X) = E(X2) - [E(X)]2
The formula means that first, we sum the square of each value times its probability then subtract the square of the mean.
For the following probability distribution
X | 1 | 2 | 3 | 4 |
P(X) | \(\frac{1}{10}\) | \(\frac{1}{5}\) | \(\frac{3}{10}\) | \(\frac{2}{5}\) |
E(X2) is equal to
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 10 Detailed Solution
Download Solution PDFConcept:
The expectation of a random variable E(X2) is given by \(\Sigma {x^2P(x) }\)
Calculation:
Given
X | 1 | 2 | 3 | 4 |
P(X) | \(\frac{1}{10}\) | \(\frac{1}{5}\) | \(\frac{3}{10}\) | \(\frac{2}{5}\) |
Expectation E(X2) is calculated by \(\Sigma {x^2P(x) }\)
⇒ E(X2) = \(\left(1^2\times\frac{1}{10}\right) +\left(2^2\times\frac{1}{5}\right)+\left(3^2\times\frac{3}{10}\right)+\left(4^2\times\frac{2}{5}\right)\)
⇒ E(X2) = \(\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}\)
⇒ E(X2) = \(\frac{100}{10}\) = 10
The value of E(X2) is equal to 10.
The correct answer is option 4.
The mean of the numbers obtained on throwing a die having written 5 on four faces, 4 on three faces, 3 on two faces and 2 on one face is?
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 11 Detailed Solution
Download Solution PDFCalculation:
Let X be the random variable representing a number on the die.
The total number of observation is ten.
P(X = 5) = \(\rm \frac{4}{10} = \frac{2}{5}\)
P(X = 4) = \(\rm \frac{3}{10} \)
P(X = 3) = \(\rm \frac{2}{10} = \frac{1}{5}\)
P(X = 2) = \(\rm \frac{1}{10} \)
Mean(X) = \(\rm \sum x_{i}p_{i}\)
Mean(X) = \(\rm 5 \times \frac{2}{5} + 4 \times \frac{3}{10} + 3 \times \frac{1}{5} + 2 \times \frac{1}{10} \)
Mean(X) = \(2 + \rm \frac{20}{10} =4 \)
In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes is respectively given by
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 12 Detailed Solution
Download Solution PDFConcept
When a fair dice is rolled, probability of getting any number is \(\frac{1}{6}\).
Suppose an experiment is repeated n times. Probability of success be p.
If X denote the number of success, then
P(X = k) = C(n, k) × pk (1 – p)n - k , 0 ≤ k ≤ n
where
\({\bf{C}}\left( {{\bf{n}},\;{\bf{k}}} \right) = \;\frac{{{\bf{n}}!}}{{{\bf{k}}!\left( {{\bf{n}} - {\bf{k}}} \right)!}}\)
Mean of X = np
Standard deviation of X \(= \sqrt {{\bf{np}}\left( {1 - {\bf{p}}} \right)}\)
Calculation
A dice is rolled 8 times so, n = 8.
In a single roll of dice, probability of getting 5 or 6 is
= probability of getting 5 + probability of getting 6
\(= \frac{1}{6} + \frac{1}{6}\)
\(= \frac{1}{3}\)
So probability of success \({\bf{p}} = \frac{1}{3}\)
\({\bf{Mean}} = {\bf{n}} \times {\bf{p}} = \frac{8}{3}\)
\({\bf{Standard}}\;{\bf{deviation}} = \sqrt {{\bf{np}}\left( {1 - {\bf{p}}} \right)} = \frac{4}{3}\)
Correct option is (2).
The variance σ2 of a random variable X is given by
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 13 Detailed Solution
Download Solution PDFConcept:
Mean = E[X]
\(E\left[ X \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } x.{f_x}\left( x \right)dx\)
\(E\left[ {{X^2}} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {x^2}{f_x}\left( x \right)dx\)
Where fx(x) is the probability density function.
Analysis:
\(Var\left( X \right) = E\left[ {{{\left( {X - {\mu _x}} \right)}^2}} \right]\)
\( = E\left[ {{X^2} - 2{\mu _x}X + \mu _x^2} \right]\)
\( = E\left[ {{X^2}} \right] - 2E\left[ {{\mu _x}X} \right] + E\left[ {\mu _x^2} \right]\)
\( = E\left[ {{X^2}} \right] - 2{\mu _x}E\left[ X \right] + \mu _x^2\)
\( = E\left[ {{X^2}} \right] - 2\mu _x^2 + \mu _x^2\)
\( = E\left[ {{X^2}} \right] - \mu _x^2\)
Since, \( E[X]= \mu _x\)
\(Var\left[ X \right] = E\left[ {{X^2}} \right] - {E^2}\left[ X \right]\)
For the following probability distribution:
X | –4 | –3 | –2 | –1 | 0 |
P(X) | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
E(X) is equal to:
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 14 Detailed Solution
Download Solution PDFConcept:
The expectation of a random variable E(X) is given by \(\Sigma {xP(x) }\)
Calculation:
Given
X | –4 | –3 | –2 | –1 | 0 |
P(X) | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
Expectation E(X) is calculated by \(\Sigma {xP(x) }\)
⇒ E(X) = (- 4) × 0.1 + (- 3) × 0.2 + (- 2) × 0.3 + (- 1) × 0.2 + 0
⇒ E(X) = - 0.4 - 0.6 - 0.6 - 0.2
⇒ E(X) = - 1.8
The value of E(X) is -1.8.
The correct answer is option 4.
A r.v.X ∼ B (n, p). If values of mean and variance of X are 18 and 12 respectively then total number of possible values of X are
Answer (Detailed Solution Below)
Mean and Variance of Random variables Question 15 Detailed Solution
Download Solution PDFConcept:
In B (n, p). n is number of observations.
mean = np
variance = npq
Calculations:
Given, a r.v.X ∼ B (n, p). n is a number of observations.
If values of mean and variance of X are 18 and 12 respectively
mean = np = 18 ....(1)
variance = npq = 12
⇒\(\rm \dfrac{npq}{np}= \dfrac{12}{18}\)
⇒ q = \(\rm \dfrac 2 3\)
we know that p = 1 - q
⇒ p = 1 - \(\rm \dfrac 2 3\)
⇒ p = \(\rm \dfrac 1 3\)
Equation (1), we have
⇒n\(\rm \times\dfrac 1 3\) = 18
⇒n = 54
⇒n = 0, 1, 2, ....54
Hence, the number of observations = 55