Mathematics Test - 29

Concept:

We know Binomial distribution is:

\({\left( {q + p} \right)^n} = \sum {n_{{C_r}}}{q^{n - r}}{p^r}\)

where p + q = 1

p is the probability of getting success and q is the probability of failure

• Mean of Binomial distribution is np
• Variance is npq.
• Standard deviation is given by the square root of the variance, as:

          \(S.D.=\;\sqrt {Variance} = \sqrt {npq} \) 

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) = nCk pk q(n - k) 

Where q = p - 1

n is the number of observations,

p is the probability of success & q is the probability of failure.

Note: The mean of a binomial distribution is np and variance is npq.

Calculation:

Given:

n = 10, p = 4/5, q = 1/5

As we know,

The mean of a binomial distribution = np

\(\rm ⇒Mean = 10 \times \frac{4}{5}=8\)

Hence, the correct option is 2.

Calculation

We know expected value of random variable

f(x) = ∑x_ip_i

where I = 1, 2, 3,-----n

f(x) = P₁x₁ + P₂x₂ + P₃x₃ + P₄x₄ + P₅x₅

⇒ f(x) = 0.1 × 2 + 0.3 × 4 + 0.4 × 6 + 0.1 × 8 + 0.1 × 10

⇒ 0.2 + 1.2 + 2.4 + 0.8 + 1

∴ f(x) = 5.6

Concept: 

To analyze the Random Variable ‘x’ two functions are used.

1) PDF (Probability Distribution Function)

2) pdf (Probability Density Function)

CDF and pdf are related as:

\(CDF = \mathop \smallint \limits_{ - \infty }^\infty PDFdx\)  ----(1)

Properties of a valid PDF:

1) \({f_X}\left( x \right) \ge 0,\;\forall \;x\;\epsilon\;R\)

∴ CDF will be bounded between 0 and 1

2) \(\mathop \smallint \limits_{ - \infty }^\infty {f_X}\left( x \right)dx = {P_X}\left( \infty \right) = 1\)      ----(2)

Here PX is CDF

\({P_X}\left( \infty \right) = \begin{array}{*{20}{c}} {{\rm{lim}}}\\ {x \to \infty } \end{array}{P_X}\left( x \right)\)

∴ CDF will always be a monotonically increasing function as the probability is always greater than or equal to 0. 

Calculation:

Given:

\(\rm PDF \ = \ f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{c}{{\sqrt x }},}&{0 < x < 4}\\ {0,}&{otherwise} \end{array}} \right.\)

Using Equation  (2):

\(\mathop \smallint \limits_{ 0 }^4 {\frac{c}{\sqrt{x}}}dx=1\)

\([2c\sqrt{x}]^{4}_{0}=1 \)

4c = 1

\(c=\frac{1}{4}\)

Hence option (3) is the correct answer.

Formula:

E(X) = μx = μ

⇒ μ = ∑ xifi = ∑xiP(X = xj)

Calculation

E(X) = μx = μ

⇒ μ = ∑ xifi = ∑xiP(X = xj)

⇒ 1 × 1/6 + 2 × 1/6 + 3 × 1/6 + 4 × 1/6 + 5 × 1/6 + 6 × 1/6

⇒ (1/6)(1 + 2 + 3 + 4 + 5 + 6)

⇒ 7/2

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:

\({\rm{P\;}}\left( {{\rm{X}} = {\rm{k}}} \right) = \left( {\begin{array}{*{20}{c}} {\rm{n}}\\ {\rm{k}} \end{array}} \right){\rm{\;}}{{\rm{p}}^{\rm{k}}}{\rm{\;}}{\left( {1 - {\rm{p}}} \right)^{{\rm{n}} - {\rm{k}}}}\)

Where, n is number of observations, p is the probability of success.

Formulas used:

• Calculate combination:

\(\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right) = \frac{{n!}}{{\left( {n - k} \right)!k!}}\)

• Factorial:

\(n! = 1 \times 2 \times 3 \times \ldots . \times \left( {n - 1} \right) \times n\)

Calculation:

Given: \(16\;P\;\left( {X = 4} \right) = P\;\left( {X = 2} \right)\)

To find the value of p

Calculating probabilities of random variable X,

\(P\;\left( {X = 4} \right) = \left( {\begin{array}{*{20}{c}} 6\\ 4 \end{array}} \right)\;{p^4}\;{\left( {1 - p} \right)^2}\)

\(P\;\left( {X = 2} \right) = \left( {\begin{array}{*{20}{c}} 6\\ 2 \end{array}} \right)\;{p^2}\;{\left( {1 - p} \right)^4}\)

Given,

\(16{\rm{\;P\;}}\left( {{\rm{X}} = 4} \right) = {\rm{P\;}}\left( {{\rm{X}} = 2} \right)\)

\( \Rightarrow 16\left( {\begin{array}{*{20}{c}} 6\\ 4 \end{array}} \right)\;{p^4}\;{\left( {1 - p} \right)^2} = \left( {\begin{array}{*{20}{c}} 6\\ 2 \end{array}} \right)\;{p^2}\;{\left( {1 - p} \right)^4}\)

\(\Rightarrow 16 \times \frac{{6!}}{{2!4!}}\;{p^4}\;{\left( {1 - p} \right)^2} = \frac{{6!}}{{2!4!}}\;{p^2}\;{\left( {1 - p} \right)^4}\)

\(\Rightarrow 16\;{p^2} = {\left( {1 - p} \right)^2}\)

\(\Rightarrow \pm 4p = 1 - p\)

\( \Rightarrow p = \frac{1}{5}\) or \(p = - \frac{1}{3}\)

Hence,\({\rm{p}} = \frac{1}{5}\)

Concept:

• \(P(X=r)=n_{c_{r}}(p)^{r}(q)^{n-r}\)
• Required probability = 1 - Not required probability

Calculation:

Here number of attempt (n) = 5

p = 1/2 and q = 1/2

P(X is greater than equal to 2) = 1 - P (X < 2) = 1 - P(X = 0) - P(X = 1)

∴ Required probability = 1 - P(X = 0) - P(X = 1)

 \(\begin{array}{l} \quad 1-{ }^{5} C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{5}-{ }^{5} C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{4} \\ \Rightarrow 1-\left(\frac{1}{2}\right)^{5}-5\left(\frac{1}{2}\right)^{5} \\ \Rightarrow 1-\left(\frac{1}{2}\right)^{5}[1+5] \\ \Rightarrow 1-\frac{6}{32} \\ \Rightarrow \frac{13}{16} \end{array}\)

Hence, option (4) is correct. 

Concept:

The mean value of (μ) of the probability distribution of a variate X is commonly known as expectation and it is denoted by E[X].

If f(x) is the probability density function of the variate X, then

Discrete distribution: \(E\left[ X \right] = \mathop \sum \limits_i {x_i}f\left( {{x_i}} \right)\)

Continuous distribution: \(E\left( X \right) = \mathop \smallint \limits_{ - \infty }^\infty xf\left( x \right)dx\)

Calculation:

\(E\left( {{X^2}} \right) = \mathop \smallint \limits_0^1 {x^2}\left( {3{x^2}} \right)dx\)

Concept:

For any random variable x the variance of x is the expected value of the squared difference between x and its expected value, i.e.

Var(x) = E(x²) - (E(x))²

E(x2) = Var(x) + (E(x))2

Where, E(x2) = Expected squared value; E(x) = Expected value;

Calculation:

Given:

Var(x) = 1; E(x) = 5;

E(x²) = Var(x) + (E(x))² = 1 + (5)2

E(x2) = 26

Concept:

• Mean = np
• Variance = npq 
• \(P(X=r)=n_{c_{r}}(p)^{r}(q)^{n-r}\)

Calculation:

Here, mean = np = 2 and 

Variance = npq = 1

(npq)/(np) = 1/2

⇒ q = 1/2

Now, p + q = 1

∴ p = 1/2

n(1/2) = 2   (∵ np = 2)

⇒ n = 4

\(P(X=r)=n_{c_{r}}(p)^{r}(q)^{n-r}\)

\(\begin{aligned} P(x=0) &={ }^{4} C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{4} \\ &=(1)(1)\left(\frac{1}{16}\right) \\ &=\left(\frac{1}{16}\right) \end{aligned}\)

Hence, option (4) is correct.