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, \( \mu = 200 \)

Variance of the binomial distribution, \( \sigma^{2} = 160 \)

For a binomial random variable,

\( \mu = np \quad\text{and}\quad \sigma^{2} = np(1-p). \)

From the mean,

\( np = 200 \;\Longrightarrow\; p = \dfrac{200}{n}. \)

Using the variance,

\( np(1-p) = 160 \;\Longrightarrow\; n\bigl(\tfrac{200}{n}\bigr)\!\Bigl(1-\tfrac{200}{n}\Bigr)=160 \;\Longrightarrow\; 200\Bigl(1-\tfrac{200}{n}\Bigr)=160. \)

Simplifying,

\( 1-\tfrac{200}{n} = 0.8 \;\Longrightarrow\; \tfrac{200}{n} = 0.2 \;\Longrightarrow\; n = \dfrac{200}{0.2} = 1000. \)

∴ The number of trials is \( n = 1000 \).

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 = \(\frac{15}{2}\)

⇒ np + npq = \(\frac{15}{2}\)

⇒ np(1 + q) = \(\frac{15}{2}\)

⇒ 10p(1 + 1 - p) = \(\frac{15}{2}\)

⇒ 10p(2 - p) = \(\frac{15}{2}\)

⇒ 20p - 10p² = \(\frac{15}{2}\)

⇒ 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 = \(\frac{1}{2}\) or p = \(\frac{3}{2}\)

Since p cannot be greater than 1, p = \(\frac{1}{2}\)

⇒ q = 1 - p = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)

Variance = npq = 10 × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\) = 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) = \(\sum xP(X = x)\) = 0 × \(\frac{1}{8}\) + 1 × \(\frac{3}{8}\) + 2 × \(\frac{3}{8}\) + 3 × \(\frac{1}{8}\) = \(\frac{3}{2}\)

Also, E(X²) = \(\sum x^2P(X = x)\) = 0 × \(\frac{1}{8}\) + 1 × \(\frac{3}{8}\) + 4 × \(\frac{3}{8}\) + 9 × \(\frac{1}{8}\) = 3

∴ Var(X) = E(X2) - [E(X)]² = 3 - \(\frac{9}{4}\) = 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 = \(\frac{3}{2}\)

∴ np = 3 and npq = \(\frac{3}{2}\)

⇒ q = \(\frac{1}{2}\)

⇒ p = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)

Substituting the value of p in mean,

\(n\times\frac{1}{2}=3\)

⇒ 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 = \(\sqrt{npq}\)

∴ Standard deviation of a binomial distribution is given by, σ = \(\sqrt{npq}\)

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) = \(\rm ^{16}C_1 \left(\frac{1}{2}\right)^1 \times \left(\frac{1}{2}\right)^{16-1}\)

\(\Rightarrow16 \times \frac{1}{2^{16}}\\\Rightarrow2^4 \times \frac{1}{2^{16}}\\\therefore\frac{1}{2^{12}}\)

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)

⇒ 1 - P(X + Y = 0)

∴ 1 - (2/3)⁸

Concept:

• Mean = μ = np
• \({\rm{Standard\;deviation}} = {\rm{\sigma }} = \sqrt {{\rm{npq}}} {\rm{\;}}\)

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

Calculation:

Given: Mean = μ = np = 12

\({\rm{Standard\;deviation}} = {\rm{\sigma }} = \sqrt {{\rm{npq}}} = 2\)

∴ 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 = \(2\over3\) and variance = npq = \(5\over9\)

\(\rm variance \over mean\) = \({5\over9}\over{2\over3}\)

q = \(5\over6\)

p = 1 - q = \(1\over6\)

np = \(2\over3\)

n = 4

∴ The probability that random variable D = 2

P = \(\rm {^nC_2}p^2q^{n-2}\)

P = \(\rm {^4C_2}\times\left({1\over6}\right)^2\times\left({5\over6}\right)^2\)

P = \(\rm 6\times{1\over36}\times{25\over36}\)

P = \(25\over216\)

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 = \(\rm \sqrt{Var(X)}\).

Calculation:

We have x₁ = 0, x₂ = 1, x₃ = 2 and p₁ = \(\dfrac{25}{36}\), p₂ = \(\dfrac{5}{18}\), p₃ = \(\dfrac{1}{36}\).

∑pixi = \(\dfrac{25}{36}\times 0+\dfrac{5}{18}\times 1+\dfrac{1}{36}\times 2=\dfrac{6}{18}=\dfrac{1}{3}\).

∑pi(xi)2 = \(\dfrac{25}{36}\times 0^2+\dfrac{5}{18}\times 1^2+\dfrac{1}{36}\times 2^2=\dfrac{7}{18}\).

∴ Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = \(\dfrac{7}{18}-\left(\dfrac{1}{3}\right)^2=\dfrac{5}{18}\).

⇒ Standard Deviation = σ = \(\rm \sqrt{Var(X)}=\sqrt{\dfrac{5}{18}}=\dfrac{1}{3}\sqrt{\dfrac{5}{2}}\).

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 = \(\rm\dfrac {8}{10}\)

⇒ 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
• \({\rm{Standard\;deviation}} = {\rm{\sigma }} = \sqrt {{\rm{npq}}} {\rm{\;}}\)
• 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 < μ

∴ Mean > Variance

Concept:

Binomial distribution The binomial distribution formula is:P (X = r) = \({{\bf{n}}_{{{\bf{C}}_{\bf{r}}}}}{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}\)

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

• \({{\rm{n}}_{{{\rm{C}}_{\rm{r}}}}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!{\rm{r}}!}}\)

• Mean = np
• Variance = npq

Calculation:

Given: mean = 2, variance = 1

\(\frac{{{\rm{mean}}}}{{{\rm{varaince}}}} = \frac{{{\rm{np}}}}{{{\rm{npq}}}} = \frac{2}{1}\)

\(\Rightarrow {\rm{q}} = \frac{1}{2}\)

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)

\( \Rightarrow 1 - {4_{{{\rm{C}}_0}}}{\left( {\frac{1}{2}} \right)^0}{\left( {\frac{1}{2}} \right)^4} - {4_{{{\rm{C}}_1}}}{\left( {\frac{1}{2}} \right)^1}{\left( {\frac{1}{2}} \right)^3}\)    (P (X = r) = \({{\rm{n}}_{{{\rm{C}}_{\rm{r}}}}}{{\rm{p}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\))

\( \Rightarrow 1 - \left( 1 \right)\left( 1 \right){\left( {\frac{1}{2}} \right)^4} - \left( 4 \right)\left( {\frac{1}{2}} \right){\left( {\frac{1}{2}} \right)^3}\)         (∵ \({{\rm{n}}_{{{\rm{C}}_0}}} = 1{\rm{\;and\;}}{{\rm{n}}_{{{\rm{C}}_1}}} = {\rm{n}}\) )

\( \Rightarrow 1 - \left( {\frac{1}{{16}}} \right) - \frac{4}{{16}}\)

\(\Rightarrow 1 - \left( {\frac{5}{{16}}} \right)\)

\(\Rightarrow \left( {\frac{{11}}{{16}}} \right)\)

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)

\(\frac{{{\rm{np}}\left( {1 - {\rm{p}}} \right)}}{{{\rm{np}}}} = \frac{{\frac{5}{9}}}{{\frac{2}{3}}}\)

⇒ (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

\(\Rightarrow {\rm{P}}\left( {{\rm{x}} = 2} \right) = \frac{{25}}{{{6^4}}}\)

\(\Rightarrow {\rm{P}}\left( {{\rm{X}} = 2} \right) = \frac{{25}}{{216}}\)

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