Engineering Mathematics Test 10

Concept:

Poisson distribution is

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

Where m is the mean of the distribution.

For Poisson distribution: Mean = Variance = m

Calculation:

P (x = 1) = P(x = 2)

\(\frac{{{m^1}{e^{ - m}}}}{{1!}} = \frac{{{m^2}{e^{ - m}}}}{{2!}}\)

m = m²/2

The line of regression of y on x is

\(y - \bar y = r \cdot \left( {\frac{{{\sigma _y}}}{{{\sigma _x}}}} \right)\left( {x - \bar x} \right)\)

The line of regression of x on y is

\(x - \bar x = r \cdot \left( {\frac{{{\sigma _x}}}{{{\sigma _y}}}} \right)\left( {y - \bar y} \right)\)

\(r\left( {\frac{{{\sigma _y}}}{{{\sigma _x}}}} \right)\) is called the regression co-efficient of y on x and is denoted by b_yx

\(r\left( {\frac{{{\sigma _x}}}{{{\sigma _y}\;}}} \right)\) is called the regression co-efficient of x on y and is denoted by b_xy

If the line is in the form of y₁ = m₁ x₁ + c₁, then the regression co-efficient b_yx = m₁

If the line is in the form of x₂ = y₂ m₂ + c₂, then the regression co-efficient b_xy = m₂

Correlation co-efficient \(\left( r \right) = \sqrt {{b_{yx}}{b_{xy}}}\)

Let X denote the number of questions asked to Girish.

The Last one he answers wrongly.

P(X = k) = P^k-1 q, k = 1, 2, 3, ….

The probability that he will quit after answering an odd number of questions = 0.9

\( \Rightarrow \;\mathop \sum \limits_{k = 0}^\infty P\left( {X = 2k + 1} \right) = 0.9\;\)

\( \Rightarrow \;\mathop \sum \limits_{k = 0}^\infty {P^{2k}}q = 0.9\) 

⇒ (q + P²q + P⁴ q + …) = 0.9

Concept:

Poisson distribution:

A Poisson random variable describes the total number of events that happen in a certain time period.

A discrete random variable X is said to have a Poisson distribution with parameter λ (λ > 0) if the pdf of X is

\(P\left( {X = x} \right) = \frac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}\;;x = 0,\;1,\;2,\; \ldots \)

Calculation:

Twice the average incidence would be 4 cases.

Let the random variable X = number of cases in 1 million people.

Has Poisson distribution with parameter 2.

P (X ≥ 4) = 1 – P (X ≤ 3)

\( = 1 - \left( {{e^{ - 2}}\frac{{{2^0}}}{{0!}} + {e^{ - 2}}\frac{{{2^1}}}{{1!}} + {e^{ - 2}}\frac{{{2^2}}}{{2!}} + {e^{ - 3}}\frac{{{2^3}}}{{3!}}} \right) = 0.143\) 

Concept:

Poisson distribution:

A Poisson random variable describes the total number of events that happen in a certain time period.

A discrete random variable X is said to have a Poisson distribution with parameter λ (λ > 0) if the pdf of X is

\(P\left( {X = x} \right) = \frac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}\;;x = 0,\;1,\;2,\; \ldots \)

Calculation:

The average number of defects (λ) = 3

For k defects:

\(P\left( {X = k} \right) = \frac{{{e^{ - 3}}\;{3^k}}}{{k!}}\) 

Where X be the number of defects per wafer.

The redundancy will not be sufficient when X > 4

P(X > 4) = 1 – P(X = 0) – P(X = 1) – P(X = 2) – P(X = 3) – P(X = 4)

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

= 0.1847

X and Y are distributed uniformly over the interval [0, 1].

The probability that the problem get solved in 20 minutes is 1/3 for both X and Y.

Case (i): Only Sachin solved in less than 20 minutes

\({P_1} = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}\) 

Case (ii): Only Saurabh solved in less than 20 minutes

\({P_2} = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\)

Case (iii): Both Sachin and Saurabh solved in less than 20 minutes

\({P_3} = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\) 

Total probability that the problem will be solved in less than 20 minutes is,

\(P = \frac{2}{9} + \frac{2}{9} + \frac{1}{9} = \frac{5}{9}\) 

Concept:

The line of regression of y on x is

\(y - \bar y = r \cdot \left( {\frac{{{\sigma _y}}}{{{\sigma _x}}}} \right)\left( {x - \bar x} \right)\)

The line of regression of x on y is

\(x - \bar x = r \cdot \left( {\frac{{{\sigma _x}}}{{{\sigma _y}}}} \right)\left( {y - \bar y} \right)\)

\(r\left( {\frac{{{\sigma _y}}}{{{\sigma _x}}}} \right)\) is called the regression co-efficient of y on x and is denoted by b_yx

\(r\left( {\frac{{{\sigma _x}}}{{{\sigma _y}\;}}} \right)\) is called the regression co-efficient of x on y and is denoted by b_xy

If the line is in the form of y₁ = m₁ x₁ + c₁, then the regression co-efficient b_yx = m₁

If the line is in the form of x₂ = y₂ m₂ + c₃, then the regression co-efficient b_xy = m₂

Correlation co-efficient \(\left( r \right) = \sqrt {{b_{yx}}{b_{xy}}}\)

Calculation:

Given equation are

y = 0.516 x + 33.73

x = 0.512 y + 32.52

From the above equations,

b_yx = 0.516, b_xy = 0.512

Correlation co-efficient \(\left( r \right) = \sqrt {{b_{yx}}{b_{xy}}}\)

Let X be the number of successful hits.

Suppose n missiles are fired.

P(X ≥ 3) ≥ 0.95

From binominal distribution,

1 – P(X < 3) ≥ 0.95

P(X < 3) ≤ 0.05

⇒ P(X = 0) + P(X = 1) + P(X = 2) ≤ 0.05

\( \Rightarrow {n_{{c_0}}}{p^0}{q^n} + {n_{{c_1}}}p\;{q^{n - 1}} + {n_{{c_2}}}{p^2}{q^{n - 2}} \le 0.05\) 

\( \Rightarrow {\left( {\frac{1}{4}} \right)^n} + n \cdot \left( {\frac{3}{4}} \right){\left( {\frac{1}{4}} \right)^{n - 1}} + \left( {\frac{n}{2}} \right){\left( {\frac{3}{4}} \right)^2}{\left( {\frac{1}{4}} \right)^{n - 2}} \le 0.05\) 

⇒ 10 (9n² – 3n + 2) ≤ 4ⁿ.