Engineering Mathematics Test 8

For Poisson distribution

\(P\left( r \right) = \frac{{{e^{ - m}}{m^r}}}{{r!}}\)

Given that

2P(X = 2) = P(X = 1) + 2P(X = 0)

\( \Rightarrow 2\left[ {\frac{{{e^{ - m}}{m^2}}}{{2!}}} \right] = \left[ {\frac{{{e^{ - m}}{m^1}}}{{1!}}} \right] + 2\left[ {\frac{{{e^{ - m}}{m^0}}}{{0!}}} \right]\)

⇒ m² = m¹ + 2

⇒ m² – m – 2 = 0

⇒ m² – 2m + m – 2 = 0

⇒ m(m - 2) + 1(m - 2) =

⇒ m = 2, -1

In Poisson distribution

Variance = mean = m = 2

\({q^2} + 2pq + {p^2} = 1\)

\({\left( {p + q} \right)^2} = 1\)

\(\therefore \left( {p + q} \right) = 1\)

\(mean = \mu = \mathop \sum \limits_{i = 0}^3 {p_i}{x_i}\)

\(= 0 \times {q^2} + 1 \times 2pq + 2 \times {p^2}\)

\(= 2p\left( {q + p} \right)\)

\(\therefore \mu = 2p\)

\(\mathop \sum \limits_{i = 0}^2 x_i^2{p_i} = \;0 \times {q^2} + 1 \times 2pq + 4 \times {p^2}\)

\(\mathop \sum \limits_{i = 0}^2 x_i^2{p_i} = \;\;2pq + 4{p^2}\)

\(Var\left( x \right) = \;\mathop \sum \limits_{i = 0}^2 x_i^2{p_i} - {\mu ^2}\)

\(Var\left( x \right) = 2pq + 4{p^2} - 4{p^2}\)

Uniformly distributed in (-2, 3), on X-axis.

X ~ U(-2,3)

Comparing with X ~ U(a,b)

Sample space = {-2, -1, 0, 1, 2, 3}

Here, b= 3, a= -2

\(Variance = \frac{1}{{12}}{\left( {b - a} \right)^2}\)

\({\rm{Variance}} = {\rm{}}\frac{1}{{12}}{\left( {3 - {\rm{\;}}\left( { - 2} \right)} \right)^2}{\rm{\;}} = {\rm{\;}}\frac{{25}}{{12}}\)

Mean of the uniform distributed variable is given as

\(mean = \frac{{a + b}}{2}\)

\(\therefore mean = \frac{{-2 + 3}}{2} = \frac{1}{2}\)

Explanation:

P(2 boys and 2 girls) \(= {\rm{\;P}}\left( {{\rm{X\;}} = {\rm{\;}}2} \right) = 4{C_2}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^{4 - 2}} = 6 \times {\left( {\frac{1}{2}} \right)^4}\)

∴ P(2 boys and 2 girls) \(= {\rm{\;P}}\left( {{\rm{X\;}} = {\rm{\;}}2} \right) = \frac{3}{8}\)

Now,

Number of families having 2 boys and 2 girls = N ⋅ P(X = 2)

(where N is the total number of families considered)

∴ Number of families having 2 boys and 2 girls \( = 800 \times \frac{3}{8}\)

Concept:

Binomial distribution:

Let p is the probability that an event will happen in a single trail (called the probability of success) and

q = 1 – p is the probability that an event will fail to happen (probability of failure)

The probability that the event will happen exactly r times in n trails (i.e. x successes and n – r failures will occur) is given by the probability function

\(f\left( x \right) = P\left( {X = r} \right) = {n_{{C_r}}}{p^r}{q^{n - r}}\)

where the random variable X denotes the number of successes in n trials and r = 0, 1, 2, … n

For Binomial distribution,

Mean = μ = np

Variance = σ2 = npq

Standard deviation = σ = √(npq)

Calculation:

Let V be the event that occurs in a trial with probability p. Mathematical expectation E of the number of trials to first occurrence of V in a sequence of trials is E = 1/p.

Similarly, the expected number of trials to get n^th success = n/p

Given that, the probability (p) = 1/5 = 0.2

The expected number of trials to get first success = 1/0.2 = 5

So, the expected number of failures preceding the first success is 4

The expected number of trials to get second success = 2/0.2 = 10

Expected number of successes in first 50 trials = np = 50 × 0.2 = 10

\(\frac{H}{D} = \frac{4}{3}\;\)

\(4x + 3x = 1\)

\(\therefore x = \frac{1}{7}\)

\(p\left( H \right) = \frac{4}{7}\;,\;p\left( D \right) = \frac{3}{7}\)

\(p\left( {D \ge 4} \right) = p\left( {D = 4} \right) + p\left( {D = 5} \right)\;\)

\(p\left( {D \ge 4} \right)\; = \;\left( {\begin{array}{*{20}{c}}5\\4\end{array}} \right)\;{\left( {\frac{3}{7}} \right)^4}\;{\left( {\frac{4}{7}} \right)^1}\; + \;\left( {\begin{array}{*{20}{c}}5\\5\end{array}} \right)\;{\left( {\frac{3}{7}} \right)^5}\;{\left( {\frac{4}{7}} \right)^0}\)

\(p(D≥4)=0.11 \)

Data:

Mean = \(\lambda \) = 3.

Formula:

\({\rm{p}}\left( {{\rm{X}} = {\rm{x}},{\rm{\lambda }}} \right) = \frac{{{{\rm{\lambda }}^{\rm{x}}}}}{{{\rm{x}}!{{\rm{e}}^{\rm{\lambda }}}}}\) 

Calculation:

\(p\left( {X \le 3,\;\lambda = 3} \right)\)

\( = p\left( {X = 0,\;\lambda = 3} \right) + \;p\left( {X = 1,\;\lambda = 3} \right) + p\left( {X = 2,\;\lambda = 3} \right)\)

\( = \frac{{{3^0}}}{{0!{e^3}}} + \frac{{{3^1}}}{{1!{e^3}}} + \frac{{{3^2}}}{{2!{e^3}}}\)

\(= \frac{1}{{{e^3}}} + \frac{3}{{{e^3}}} + \frac{9}{{2{e^3}}}\)

\(= \frac{{17}}{{2{e^3}}}\)

Concept: 

If f(x) is probability density function, then

1. \(\mathop \smallint \nolimits_{ - \infty }^\infty f\left( x \right)dx = 1\)

2. Mean \(= E\left( x \right) = \mathop \smallint \nolimits_{ - \infty }^\infty x \cdot f\left( x \right)dx\)

Calculation:

X has a probability density function f(x)

\(\mathop \smallint \limits_0^\infty {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 1\)

\(\Rightarrow \mathop \smallint \limits_0^\infty {\rm{k\;}}{{\rm{x}}^{\rm{n}}}{{\rm{e}}^{ - {\rm{x}}}}{\rm{dx}} = 1\)

We know that,

\({\rm{\Gamma m}} = \mathop \smallint \limits_0^\infty {{\rm{e}}^{ - {\rm{x}}}}{{\rm{x}}^{{\rm{m}} - 1}}{\rm{dx}}\)

\(\Rightarrow {\rm{k\Gamma }}\left( {{\rm{n}} + 1} \right) = 1\) → (1)

Given that mean = 3

\(\Rightarrow \mathop \smallint \limits_0^\infty {\rm{x\;f}}\left( x \right){\rm{dx}} = 3\)

\(\Rightarrow \mathop \smallint \limits_0^\infty {\rm{k\;}}{{\rm{x}}^{{\rm{n}} + 1}}{{\rm{e}}^{ - {\rm{x}}}}{\rm{dx}} = 3\) 

\(\Rightarrow {\rm{k\Gamma }}\left( {{\rm{n}} + 2} \right) = 3\) → (2)

From equations (1) and (2)

\(\Rightarrow \frac{{k{\rm{\Gamma }}\left( {n + 1} \right)}}{{k{\rm{\Gamma }}\left( {n + 2} \right)}} = \frac{1}{3}\)

\(\Rightarrow 3{\rm{\Gamma }}\left( {n + 1} \right) = {\rm{\Gamma }}\left( {n + 2} \right)\)

We know that \({\rm{\Gamma }}n = \left( {{\rm{n}} - 1} \right)!\)

⇒ 3n! = (n + 1)!

⇒ 3 = n + 1

⇒ n = 2

From equation (1),