Engineering Mathematics Test 9

Concept:

Uniform Distribution:

If x is continuous Random variable defined in [a, b], such that its p.d.f is given as:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\;\;\;\;\;;\;\;\;\;a \le x \le b}\\{0\;\;\;\;\;\;\;\;\;\;\;\,\,;\;\;\;\;otherwise}\end{array}} \right.\)

Then x is called uniform Random variable.

 Continuous probability Distribution:

Let x is Continuous Random Variable, having p.d.f.  f(x), then we have following results:

1) f(x) = Prob at x

2) f(x) ≥ 0

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

4) \(P\left( {a \le x \le b} \right) = \mathop \smallint \limits_a^b f\left( x \right)\)

5) \(E\left( x \right) = \mathop \smallint \limits_{ - \infty }^\infty xf\left( a \right)dx\) 

Calculation:

Let x be the length of the shorter piece of pencil. Hence it varies between \(\left[ {0 - \frac{1}{2}} \right].\)

By using uniform distribution, p.d.f of the given condition:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{b - a}}\;\;\;\;;\;\;\;a < x \le b}\\{0,\;\;\;\;\;otherwise}\end{array}} \right.\)

\(= \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{\frac{1}{2} - 0}} = 2\;\;\;\;\;\;;\;\;\;0 < x < \frac{1}{2}}\\{0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{array}} \right.\)

From the definition of continuous probability distribution,

\(E\left( x \right) = \mathop \smallint \limits_{ - \infty }^\infty x\;f\left( x \right)dx = \mathop \smallint \limits_0^{\frac{1}{2}} x \cdot 2\;dx = \frac{1}{4}\)

Concept:

(i) A random variable X is said to be of continuous type if its distribution function FX is continuous everywhere.

(ii) A random variable X with cumulative distribution function FX is said to be of absolutely continuous type if there exists an integral function fX : R → R such that fX(x) ≥ 0, for x ϵ R. It should also satisfy:

\({F_X}\left( x \right) = \mathop \smallint \limits_{ - \infty }^x {f_X}\left( x\right)dx,x\epsilon R\)

The function fX is called the probability density function (p.d.f.) of random variable X.

And FX(x) is the cumulative distribution function evaluated as the integration of the density function.

Properties of a valid PDF:

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

\(\mathop \smallint \limits_{ - \infty }^\infty {f_X}\left( x \right)dt = {F_X}\left( \infty \right) = 1\)

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

Calculation:​

​\(\begin{array}{l} \mathop \smallint \limits_ \infty ^ \infty f\left( x \right)dx = 1\\ \mathop \smallint \limits_0^2 kx\;dx + \mathop \smallint \limits_2^4 2k\;dx + \mathop \smallint \limits_4^6 \left( { - kx + 6k} \right)dx = 1 \end{array}\)

\(\begin{array}{l} \Rightarrow {\left[ {\frac{{k{x^2}}}{2}} \right]_0}^2 + {\left[ {2kx} \right]_2}^4 + {\left[ {\frac{{ - k{x^2}}}{2} + 6kx} \right]_4}^6 = 1\\ \Rightarrow \frac{k}{2}\left( 4 \right) + 2k\left( 2 \right) + \left( {\frac{{ - k}}{2}\left[ {36 - 16} \right]} \right) + 6k\left( {6 - 4} \right) = 1 \end{array}\)

⇒ 2k + 4k - 10k + 12k = 1

\(\begin{array}{l} \Rightarrow k = \frac{1}{8} \end{array}\)

Important Points:

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

Concept:

Binomial Probability Distribution:

Let X is the discrete Random variable such that its p.m.f is defined as

\(P\left( {X = r\;success} \right) = {\;^n}{C_r}{p^r}{q^{n - r}} \approx {\left( {q + p} \right)^n}\)

Where,

p = probability of success & q = probability of failure & p + q = 1

then X is said to follow Binomial distribution.

1) Parameter/statistical Attributes: Those unknowns which are necessary to evaluate complete distribution are called parameters.

2) Mean : E(x) = ∑ x_ip_i = np

3) Variance: E(x²) – (E(x))² = npq

4) Standard Deviation: \(\sqrt {npq}\)

Calculation:

Comparing \({\left( {q + p} \right)^n}\;with\;{\left( {\frac{1}{3} + \frac{2}{3}} \right)^{10}}\)

\(q = \frac{1}{3},\;p = \frac{2}{3},n = 10\)

∴ Standard Deviation \(= + \sqrt {npq}\) \(= \sqrt {\frac{1}{3} \times \frac{2}{3} \times 10} \approx 1.49\)

Variance \(= npq = \frac{1}{3} \times \frac{2}{3} \times 10 = \frac{{20}}{9}\)

∴ Required sum \(= 1.49 + \frac{{20}}{9} = 3.712\;\)

Concept:

Binomial Probability Distribution:

Let X is the discrete Random variable such that its p.m.f is defined as

\(P\left( {X = r\;success} \right) = {\;^n}{C_r}{p^r}{q^{n - r}} \approx {\left( {q + p} \right)^n}\)

Where,

p = probability of success & q = probability of failure & p + q = 1

then X is said to follow Binomial distribution.

1) Parameter/statistical Attributes: Those unknowns which are necessary to evaluate complete distribution are called parameters.

2) Mean : E(x) = ∑ x_ip_i = np

3) Variance: E(x²) – (E(x))² = npq

4) Standard Deviation: \(\sqrt {npq}\) 

Calculation:

Probability of 6^th tail in 10^th Throw = Probability of exactly 5 tails in 9 throw and probability of T in 10^th throw:

Now, by Binomial distribution:

Probability of 5T in 9 throws \({ = ^{\;9}}{C_5}\;{\left( {\frac{1}{2}} \right)^5}{\left( {\frac{1}{2}} \right)^4} = \frac{{63}}{{256}}\) 

Probability of T in 10^th throw \(= \frac{1}{2}\)

∴ Required probability \(= \frac{{63}}{{256}} \times \frac{1}{2} = \frac{{63}}{{512}}\)

For Poisson distribution

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

Given that

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

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

⇒ 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

\(f\left( x \right) = K{e^{\left( { - \frac{{{x^2}}}{{50}}\; + \;4x\; - \;200} \right)}}\)

\( = K{e^{ - \frac{1}{{50}}\;\left( {{x^2}\; - \;200x\; + \;10000} \right)}}\)

\( = K{e^{ - \frac{1}{{50}}\;{{\left( {x\; - \;100} \right)}^2}}}\)

Probability density function of normal distribution is

\(f\left( x \right) = \frac{1}{{\sigma \sqrt {2\pi } }}\;{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}\;\;\;\;\)

\({\sigma ^2}\; = \;25,\mu \;\;= \;100\;\)

\(\sigma \; = \; + \sqrt {var} \; = \;\sqrt {25} \; = \;5\)\(\)

\(\;sum\; = \;\sigma + \mu \; = \;100\; + \;5\; = \;105\)

Note:
σ = standard deviation

σ² = variance

μ = mean

Concept:

Expected Value: it is average of the probability distribution

\(E\left( x \right) = \sum {p_i}{x_i}\)

Calculation:

Let N₁ = 100 students and N₂ = 80 questions.

For a single student in a single question, expected marks obtained (x) = {Either 1 or -0.25}

Probability distribution:

X :    1      -0.25

P(x):  ¼      ¾

\(\therefore E\left( x \right) = \sum {p_i}{x_i} = 1 \times \frac{1}{4} + \left( { - 0.25} \right) \times \frac{3}{4} = \frac{1}{{16}}\)

Hence expected marks obtained by single students in 1 question \(= \frac{1}{{16}}\)

∴ Expected marks by single student in complete test \(\; = \frac{1}{{16}} \times 80 = 5\;marks\)