Mathematics Test - 9

Probability Density Function:

It indicates the distribution of total probability to various random variables (R.V)

\(f\left( x \right) = \frac{d}{{dx}}F\left( x \right)\)

f(x) : pdf (Probability density function)

F(x) : PDF (probability Distribution function)

For a continuous R.V., the probability of occurrence at a particular value of ‘x’ will be ‘zero’. [Area approaches to zero].

Since the given R.V. is continuous and the period/interval is [a, b]:

P[x = c] will be zero.

∴ P[x = 8] will be zero.

Important Notes:

Properties of pdf:

1) \(\mathop \smallint \nolimits_{ - a}^x f\left( x \right)dx = 1\) (or) Total area = 1

2) \(P\left[ {x \le {x_1}} \right] = F\left( {{x_1}} \right) = \mathop \smallint \nolimits_{ - \infty }^{{x_1}} f\left( x \right)dx\)

3) \(P\left[ {x > {x_1}} \right] = 1 - F\left( {{x_1}} \right) = \mathop \smallint \nolimits_{{x_1}}^\infty f\left( x \right)dx\) 

4) \(P\left[ {{x_1} < x \le {x_2}} \right] = \mathop \smallint \nolimits_{{x_1}}^{{x_2}} f\left( x \right)dx\)

For Discrete R.V., the Probabilities are defined at a particular value. Whereas for continuous R.V. probabilities are defined for a particular interval.

Calculation:

\(\rm \sum p_{i} = 1\)

\(\rm \frac{13}{k} + \frac{15}{k} + \frac{17}{k} + \frac{19}{k} = 1\)

\(\rm \frac{64}{k} = 1\)

k = 64

The value of k is 64

Calculation:

Given, np = 2       ....(1)

and npq = 1     ....(2)

On solving 1 and 2, we get

\(\rm p = \frac{1}{2}, q = \frac{1}{2}, n = 4\)

P(X ≥ 1)

= 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ⁴C₀  \(\rm \left (\frac{1}{2} \right )^{0} \left ( \frac{1}{2} \right )^{4}\)

= \(\rm \frac{15}{16}\)

The probability that Y takes a value greater than or equal to one is \(\rm \frac{15}{16}\)

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

Given

f(x) = { ke^-3x, x > o

          { 0, elsewhere

Concept used

\(\smallint \limits_{ - \infty }^\infty f\left( x \right)dx \) = \( \smallint \limits_{ - \infty }^0 f\left( x \right)dx\) + \(\smallint \limits_0^\infty f\left( x \right)dx\) = 1

Calculation

According to given part

⇒\(\smallint \limits_{ - \infty }^0 f\left( x \right)dx\) = 0

⇒ 0 + \( \smallint \limits_0^\infty f\left( x \right)dx\) = 0

⇒ ∫ke^-3xdx = 1

⇒ k[-e^-3x/3]

⇒ -k/3[e^-∞- e⁰] = 1

⇒ -k/3(0 – 1) = 1

⇒ k./3 = 1

∴ The value of k = 3 for PDF

Concept:

For any random variable x:

Var(x) = E(x2) - {E(x)}2 

E(x2) = Var(x) + {E(x)}2

Where, E(x) = Expected value and Var(x) = Variance.

\(Standard\;Deviation\;σ=\sqrt{Variance}\)

Calculation:

Given:

E(x) = \(\sqrt{5}\), Standard deviation σ = 4

\(Standard\;Deviation\;σ=\sqrt{Variance}\)

\(4=\sqrt{Variance}\)

Variance = 16

For any random variable x:

E(x2) = Var(x) + {E(x)}2

\(E(x^2)=16+(\sqrt{5})^2=21\)

Concept: 

E[x] = Mean of Random variable X

X = discrete random variable

The expectation is given by the formula:

\(E\left[ x \right] = \mathop \sum \limits_{i = 1}^n {x_i}\;P\left( {{x_i}} \right)\)

where xi = ith random variable and P(xi) = Probability of xi.

Calculation:

We know that,

\(E\left[ x \right] = \mathop \sum \limits_{i = 1}^n {x_i}\;P\left( {{x_i}} \right)\)

⇒ E(X) = 0(1/7) + 1(2/7) + 2 (3/7) + 3(4/7) + 4(5/7)

⇒ E(X) = 0 + 2/7 + 6/7 + 12/7 + 20/7

⇒ E(X) = 5.71

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

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

Concept:

E[x] = Mean of Random variable X

X = discrete random variable

The expectation is given by the formula:

\(E\left[ x \right] = \mathop \sum \limits_{i = 1}^n {x_i}\;P\left( {{x_i}} \right)\)

x_i = i^th random variable

P(xi) = Probability of x_i

Calculation:

The probability distribution of X is:

\(E\left( X \right) = \mathop \sum \nolimits_{n = 1}^\infty n\left( {\frac{1}{{{3^n}}}} \right)\;\)

= \(\left( {\frac{1}{3}} \right) + 2{\left( {\frac{1}{3}} \right)^2} + 3{\left( {\frac{1}{3}} \right)^3} + \ldots + \infty \)

= \(\frac{1}{3}{\left( {1 - \frac{1}{3}} \right)^{ - 2}} = \frac{3}{4}\)

Concept:

Binomial distribution:

\(b\;\left( {x;n\;,\;p} \right) = \;{\;^n}{C_x} \times {p^x} \times {q^{n - x}}\) where p is the probability of success, q is the probability of failure, n is the total no. of attempts and x is the no. of successful attempts.

Calculation:

Given: A fair die is rolled 4 times.

Let p represent the probability of getting 6 when a dice is rolled = 1 / 6

Let q represent the probability of not getting 6 when a dice is rolled

= 1 – (1 / 6) = 5 / 6

As we know that, according to binomial distribution:

\(b\;\left( {x;n\;,\;p} \right) = \;{\;^n}{C_x} \times {p^x} \times {q^{n - x}}\)

Here, n = 4, x = 2, p = 1 / 6 and q = 5 / 6.

So, the probability of getting exactly 2 sixes when a fair dice is rolled 4 times

⇒ \(b\;\left( {x;n\;,\;p} \right) = \;{\;^4}{C_2} \times {\left( {\frac{1}{6}} \right)^2} \times {\left( {\frac{5}{6}} \right)^2} = \frac{{25}}{{216}}\)