Mean and Variance of Binomial Distribution MCQ Quiz - Objective Question with Answer for Mean and Variance of Binomial Distribution - Download Free PDF

Calculation:

Given,

Mean of the binomial distribution,

Variance of the binomial distribution,

For a binomial random variable,

From the mean,

Using the variance,

Simplifying,

∴ The number of trials is .

Concept Used:

For a Binomial Distribution, mean = np and variance = npq, where p is the probability of success and q is the probability of failure.

Calculation

Given: Mean + Variance =

⇒ np + npq =

⇒ np(1 + q) =

⇒ 10p(1 + 1 - p) =

⇒ 10p(2 - p) =

⇒ 20p - 10p² =

⇒ 40p - 20p² = 15

⇒ 20p² - 40p + 15 = 0

⇒ 4p² - 8p + 3 = 0

⇒ 4p² - 6p - 2p + 3 = 0

⇒ 2p(2p - 3) - 1(2p - 3) = 0

⇒ (2p - 1)(2p - 3) = 0

⇒ p = or p =

Since p cannot be greater than 1, p =

⇒ q = 1 - p = 1 - =

Variance = npq = 10 × × = = = 2.5

∴ The variance is 2.5.

Hence option 2 is correct

Calculation:

Given, three fair coins numbered 1 and 0 are tossed simultaneously.

∴ Sample space, S = {000, 001, 010, 100, 011, 101, 110, 111}

⇒ n(S) = 8

Let X represents the sum of numbers on upper most face.

P(X = 0) = 1/8

P(X = 1) = 3/8

P(X = 2) = 3/8

P(X = 3) = 1/8

∴ E(X) =  = 0 ×  + 1 ×  + 2 ×  + 3 ×  = 

Also, E(X²) =  = 0 ×  + 1 ×  + 4 ×  + 9 ×  = 3

∴ Var(X) = E(X2) - [E(X)]² = 3 -  = 0.75

∴ The variance is 0.75.

The correct answer is Option 2.

Concept:

Binomial Distribution:

• A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times.
• The binomial is a type of distribution that has two possible outcomes (the prefix “bi” means two, or twice).
• For example, a coin toss has only two possible outcomes: heads or tails, and taking a test could have two possible outcomes: pass or fail.

Key PointsThe binomial distribution has the following properties:

• The mean of the distribution (μ) is equal to np.
• The variance (σ2) is npq.

where, n: The number of trials in the binomial experiment.

p: The probability of success on an individual trial.

q: The probability of failure on an individual trial.(q = 1 - p)

Calculation:

According to the question, mean = 3 and variance = 

∴ np = 3 and npq = 

⇒ q = 

⇒ p = 1 -  = 

Substituting the value of p in mean,

⇒ n = 6.

∴ Number of trials = 6.

The correct answer is Option 2.

Concept:

Binomial Distribution:

• A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times.
• The binomial is a type of distribution that has two possible outcomes (the prefix “bi” means two, or twice).
• For example, a coin toss has only two possible outcomes: heads or tails, and taking a test could have two possible outcomes: pass or fail.

Key PointsThe binomial distribution has the following properties:

• The mean of the distribution (μ) is equal to np.
• The variance (σ²) is npq.

where, n: The number of trials in the binomial experiment.

p: The probability of success on an individual trial.

q: The probability of failure on an individual trial.

(This is equal to q = 1 - p)

Calculation:

The variance of a binomial distribution is given by:

σ2 = npq

∴ Standard deviation, σ = 

∴ Standard deviation of a binomial distribution is given by, σ = 

The correct answer is Option 3

Concept:

Binomial distribution: 

If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P(X = k) = ⁿC_k p^k q^{(n - k)}

where n is the number of observations, p is the probability of success & q is the probability of failure.

Properties:

• Mean of the distribution (μ) = np
• The variance (σ²) = npq

Calculation:

Given: 

mean μ = np = 8      ----(1)

variance σ² = npq = 4       ----(2)

Dividing equation (2) by (1), we get

q = 1/2

As we know, p + q = 1

⇒ p = 1 - q = 1/2

Put the value of n in equation (1), we get

n = 16

Now,

P(x = 1) = 

Given

If the independent random variables X,Y are Binomially distributed that means

X ~ B(n, p)

X ~ B(3, 1/3) and Y ~ B(5, 1/3)

Here, B = Binomial distribution

n = number of observation

p = probability

Calculation

Since X and Y are independent binomial random variables with p₁ = p₂ = 1/3

By adding property of Binomial distribution

⇒ X + Y ~ B(3 + 5, 1/3)

⇒ X + Y ~B(8,1/3)

We know that, P(X + Y ≥ γ) = ⁸C_r(1/3)^r(2/3)^{8 - r}

⇒ P(X + Y ≥ 1) = 1 - P( X + Y

⇒ 1 - P(X + Y = 0)

∴ 1 - (2/3)⁸

Concept:

• Mean = μ = np

Where n = number of trials; p = probability of success; q = (1-p) = probability of failure

Calculation:

Given: Mean = μ = np = 12

∴ Variance = σ² = npq = 4

⇒ 4 = np (1−p) = 12(1−p) 

 ⇒ 4/12 = (1−p) 

⇒ (1 – p) = 1/3

⇒ p = 1 – 1/3

∴ p = 2/3

Again; μ = np = 12

∴ n = 12/p = 12 × (3/2) = 18

Concept:

For the binomial distribution of probability

• Mean = np
• Variance = npq

Where n is the total number of cases, p is probability of favorable cases and q is probability of unfavorable cases(1 - p)

Calculation:

Given mean = np =  and variance = npq = 

 = 

q = 

p = 1 - q = 

np = 

n = 4

∴ The probability that random variable D = 2

P = 

P = 

P = 

P = 

Concept:

For a random variable X = x_i with probabilities P(X = x_i) = p_i:

• Mean/Expected Value: μ = ∑p_ix_i.
• Variance: Var(X) = σ² = ∑p_i(xi)² - (∑pixi)² = ∑pi(xi)2 - μ2.
• Standard Deviation: σ = .

Calculation:

We have x₁ = 0, x₂ = 1, x₃ = 2 and p₁ = , p₂ = , p₃ = .

∑pixi = .

∑pi(xi)2 = .

∴ Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = .

⇒ Standard Deviation = σ = .

Concept:

Expected Mean:

E(x) = np  

Calculations:

Given that x ~ B ( n = 10, p)

E(x) = 8 

We know that E(x) = np

⇒ np = 8 

⇒(10)p = 8

⇒ p = 

⇒ p = 0.8

Given that x ~ B ( n = 10, p) if E(x) = 8 then the value of P is 0.8

Concept:

• Mean = μ = np
• Variance = npq

Where n = number of trials; p = probability of success; q = (1-p) = probability of failure

probability of an event lies between 0 and 1.

Calculation:

As we know, Mean = μ = np

Variance = npq 

⇒ Variance = μ × q

Probability of an event lies between 0 and 1.

Therefore 0 ≤ {p, q} ≤ 1

μ × q 

∴ Mean > Variance

Concept:

Binomial distribution The binomial distribution formula is:P (X = r) =

P(X) = probability of a success on an individual trial

p = probability of “success”

r = number of success

q = probability of “failure” OR q = 1- p

n = number of trials

• Mean = np
• Variance = npq

Calculation:

Given: mean = 2, variance = 1

We know p + q = 1

So, p = 1 - ½ = ½

mean = 2

⇒ np = 2

⇒ n = 2/p

= 4

We know P(required) = 1 – P (not required)

P (X > 1) = 1 – P (X = 0) – P (X = 1)

    (P (X = r) = )

         (∵  )

Hence, option (4) is correct

Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = ⁿC_k p^k (1 - p)^{(n - k)}

Where, n is number of observations, p is the probability of success & (1 - p) is probability of failure.

Properties:

• Mean of the distribution (μ_X) = n × p
• The variance (σ²_x) = n × p × (1 - p)
• Standard deviation (σ_x) = √{np(1 - p)}

Calculation:

Given:

Mean = n × p = 2/3    …. (1)

Variance = n × p × (1 - p) = 5/9    …. (2)

Dividing equation (1) by equation (2)

⇒ (1 - p) = 5/6

⇒ p = 1/6

Put p = 1/6 in equation (1)

n × (1/6) = 2/3

⇒ n = 4

Now, P(X = 2) = ⁴C₂ × (1/6)² × (1 - 1/6)^{(4 - 2)}

P(X = 2) = 6 × (1/6)² × (5/6)²

Given

n = 25

p = 0.2

q = 1 – p = 0.8

Formula

The skewness of binomial distribution is β₁

β₁ = (q – p)/√(n × p × q)

p = probability of success event

q = probability of failure event

Calculation

β₁ = (0.8 – 0.2)/√(25 × 0.8 × 0.2)

⇒ 0.6/√(40

∴ The skewness of variable X is 0.3