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Questions Related to business maths

A examinations consists of $8$ questions in each of which  one of the $5$ alternatives is the correct one. On the assumption that a candidate who has done no preparatory  work, chooses for each questions any one of the five alternatives with equal probability, then the probability that he gets more than one correct answer is equal to:

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. ${\left( {0.8} \right)^8}$

  2. $3{\left( {0.8} \right)^8}$

  3. $1-{\left( {0.8} \right)^8}$

  4. $1-3{\left( {0.8} \right)^8}$

Show answer
Correct Option: D
Explanation:
Probability of an answers to be correct $=\dfrac{1}{5}=0.2$

Probability of an answers not to be correct $=1-0.2$
$=0.8$

Probability $\left(more\ than\ 1\ correct\right)=1-P\left(0\ correct\right)-P\left(1\ correct\right)$

$=1-^{8}{C} _{0}{\left(0.8\right)}^{8}-^{8}{C} _{1}\left(0.2\right) \left(0.8\right)$

$=1-{\left(0.8\right)}^{8}-1.6{\left(0.8\right)}^{7}$

$=1-{\left(0.8\right)}^{8}-2{\left(0.8\right)}^{8}$

$=1-3{\left(0.8\right)}^{8}$

$D$ is coorect.

There are twenty bags each containing 10 bulbs and it is knows that no bag contains more than 5 defective bulbs and 3 bags have 5 defective bulbs. 4 bags have atleast 4 defective bulbs, 5 bags have atleast 3 defective bulbs, 6 bags have atleast 2 defective bulbs and 7 bags have atleast 1 defective bulb. Then the ratio of total defective bulbs is to non-defective bulbs is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\dfrac { 4 }{ 7 } $

  2. $\dfrac { 3 }{ 7 } $

  3. $\dfrac { 2 }{ 7 } $

  4. $\dfrac { 1 }{ 7 } $

Show answer
Correct Option: A

For a Poission distribution which pair has same value.

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. (Mean, Std. Deviation)

  2. (Variance, Standard Deviation)

  3. (Mean, Variance)

  4. None of these

Show answer
Correct Option: C
Explanation:
For Poission distribution we have $\displaystyle P\left ( r \right )=\dfrac{e^{-\pi }\lambda ^{r}}{r^{i}};\left ( r=0, 1, 2,...\infty  \right)$ 
$\displaystyle$ mean $\displaystyle =\dfrac{\sum f _{i}x _{i}}{\sum f _{i}}=\sum _{r=0}^{\infty }rP\left ( r \right ),\sum f _{i}P\left ( r \right )=1$$\displaystyle =0+\lambda e^{-\lambda }+2\dfrac{\lambda ^{2}e^{-\lambda }}{2!}+3\dfrac{\lambda ^{3}e^{-\lambda }}{3!}+.....\infty$ $\displaystyle =\lambda e^{-\lambda }\left ( 1+\dfrac{\lambda }{11}+\dfrac{\lambda ^{2}}{2!}+\dfrac{\lambda ^{3}}{3!}+.....\infty  \right )$$\displaystyle =\lambda e^{-\lambda }.e^{\lambda }=\lambda $
Similarly, $\displaystyle o^{2}=$ Variance $=\sum _{r=0}^{\infty }r^{2}P\left ( r \right )-\left ( \sum _{r=0}^{\infty } rP\left ( r \right ) \right )^{2}$ $=\displaystyle \lambda e^{\lambda }\left ( e^{-\lambda }+\lambda e^{-\lambda } \right )-\lambda ^{2}=\lambda =$ mean 
$\therefore$ mean$=$variance

For a poission distribution variable $X$ is such that $P(X = 2) = 9 P(X= 4) + 90 P(X= 6)$ the mean is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $2$

  2. $3$

  3. $1$

  4. None of these

Show answer
Correct Option: C
Explanation:
For $P.D. P(X=r)=\displaystyle \dfrac{e^{-\lambda}\lambda ^{r}}{r!},r=0,1,2,\cdots$
$\therefore \displaystyle \dfrac{e^{-\lambda}\lambda ^{2}}{2!}=\dfrac{9e^{-\lambda}\lambda ^{4}}{4!}+90 \dfrac{e^{-\lambda}\lambda ^{6}}{6!}$        (given)
$\Rightarrow \displaystyle\dfrac{\lambda ^{2}}{2}=\dfrac{9}{24}\lambda ^{4}+\dfrac{90}{720}\lambda ^{6}$ 
$\Rightarrow \lambda ^{4}+3\lambda ^{2}-4=0$
$\Rightarrow \left ( \lambda^{2}+4 \right )\left ( \lambda^{2}-1 \right )=0=>\lambda= \pm 1$
$\Rightarrow \lambda=1$ as $\lambda>0$ and $(\lambda^{2}+4=0$ impossible$)$
$\therefore$ mean$=\lambda=1$

For a Poission distribution, which of the following is true

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $Mean = Mode$

  2. $Median = S.D.$

  3. $Mean = Variance$

  4. $Median = Variance$

Show answer
Correct Option: C
Explanation:
For Poission distribution we have $\displaystyle P\left ( r \right )=\dfrac{e^{-\pi }\lambda ^{r}}{r^{i}}\left ( r=0, 1, 2,...\infty  \right)$ 
$\displaystyle$ mean $\displaystyle =\dfrac{\sum f _{i}x _{i}}{\sum f _{i}}=\sum _{r=0}^{\infty }rP\left ( r \right ),\sum f _{i}P\left ( r\right )=1$$\displaystyle =0+\lambda e^{-\lambda }+2\dfrac{\lambda ^{2}e^{-\lambda }}{2!}+3\dfrac{\lambda ^{3}e^{-\lambda }}{3!}+.....\infty$ $\displaystyle =\lambda e^{-\lambda }\left ( 1+\dfrac{\lambda }{11}+\dfrac{\lambda ^{2}}{2!}+\dfrac{\lambda ^{3}}{3!}+.....\infty  \right )$$\displaystyle =\lambda e^{\lambda }.e^{\lambda }=\lambda $
similarly, $\displaystyle o^{2}=$ Variance $=\sum _{r=0}^{\infty }r^{2}P\left ( r \right )-\left ( \sum _{r=0}^{\infty } rP\left ( r \right ) \right )^{2}$ $=\displaystyle \lambda e^{\lambda }\left ( e^{-\lambda }+\lambda e^{-\lambda } \right )-\lambda ^{2}=\lambda =$ mean 
$\therefore \text{mean}=\text{variance}$

At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with a average of 5 phone calls during IO-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\displaystyle \frac{6}{5^{e}}$

  2. $\displaystyle \frac{5}{6}$

  3. $\displaystyle \frac{6}{55}$

  4. $\displaystyle \frac{6}{e^{5}}$

Show answer
Correct Option: D
Explanation:

According to Poisson distribution
$\displaystyle P(X=r)=\frac{e^{-m}m^{r}}{r!}$
$\displaystyle \therefore P(X\leq 1)=P(X=0)+P(X=1)$
$\displaystyle =e^{-m}+\frac{e^{-m} m}{1!}$
Given m $\displaystyle =$mean$ = 5 $
$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$

The probability of r successes in case of poissons distrbution is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\dfrac{e^{\gamma }m}{\angle \gamma }$

  2. $\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$

  3. $\dfrac{e^{m}\gamma }{\angle \gamma }$

  4. $\dfrac{e^{-m}m^{r}}{\angle \gamma }$

Show answer
Correct Option: D
Explanation:

in Poisson distribution, probability $ P(x; m  )=  \dfrac { { e }^{ - m  }{ m }^{ x } }{ \angle x } $
$m$ = mean  ;   $x$= number of success     $ \angle x =$ factorial $ x $
 For $r$ success  $x= r$;        
  $ P(r; m )=  \dfrac { { e }^{ -m  }{ m  }^{ r } }{ \angle r } $ 

A random variable $X$ has Poisson distribution with mean $2$. Then $P(X > 1.5)$ equals

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $2/e^{2}$

  2. $0$

  3. $1-\dfrac{3}{e^{2}}$

  4. $\dfrac{3}{e^{2}}$

Show answer
Correct Option: C
Explanation:

For Poisson distribution of mean = $\mu = 2 $ 
 P$(x ; \mu) = \frac { { e }^{ -\mu  }{ \mu  }^{ x } }{ x! }$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)]$ 

P$(0; 2) = \dfrac { { e }^{-2}{2}^{0}}{0!}={e}^{-2}$
 P$(1; 2) = \dfrac { { e }^{-2}{2}^{} } {1!}=2{e}^{-2}$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)] $
$ = 1 - [{e}^{-2}+ 2{e}^{-2} ] = 1 - 3  {e}^{-2} = 1 - \frac { 3 }{ { e }^{ 2 } } $

If $X$ is a random poisson variate such that $\alpha =p(X=1)=p(X=2)$, then $p(X=4)=$

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $2\alpha $

  2. $\dfrac{\alpha }{3}$

  3. $\alpha e^{-2}$

  4. $\alpha e^{2}$

Show answer
Correct Option: B
Explanation:

In  Poisson distribution such that  $ \alpha = p(X=1)=p(X=2) $ 
        $   p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $  
       => $ \alpha = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}  $       =>   $ \alpha =  { e }^{-\mu}{\mu}  $
                 =>  $ { e }^{-\mu}{\mu} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2} $
                 =>   $ \mu = 2 $ 
  $   p(4; \mu) = \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} =  \dfrac { { e }^{ -\mu  }{ \mu \times { \mu  }^{ 3 } } }{ 24 }  = \dfrac { \alpha {\times { 2 }^{ 3 } } }{ 24 }  = \dfrac { \alpha  }{ 3 } $  

The variance of P.D. with parameter $\lambda $ is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\lambda $

  2. $\sqrt{\lambda }$

  3. $\dfrac{1}{\lambda}$

  4. $\dfrac{1}{\sqrt {\lambda}}$

Show answer
Correct Option: A
Explanation:

$E[{ X }^{ 2 }]= \sum _{ k=0 }^{ \infty  }{ k^{ 2 } } \sum _{  }^{  }{  } \dfrac { 1 }{ k! } \lambda ^{ k }e^{ -\lambda  }\ \ = ^{ k }e^{ -\lambda  }\sum _{ 1 }^{ \infty  }{ k\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } \= ^{ k }e^{ -\lambda  }(\sum _{ 1 }^{ \infty  }{ (k-1)\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } ) +\sum _{ 1 }^{ \infty  }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ =^{ k }e^{ -\lambda  }(\sum _{ 2 }^{ \infty  }{ (\lambda \dfrac { 1 }{ (k2)! } \lambda ^{ k-2 } } )+ \sum _{ 1 }^{ \infty  }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ = ^{ k }e^{ -\lambda  }(\sum _{ i=0 }^{ \infty  }{ (\lambda \dfrac { 1 }{ i! } \lambda ^{ i } } )+ \sum _{ j=0 }^{ \infty  }{ \dfrac { 1 }{ j! } \lambda ^{ j } } )\ = ^{ k }e^{ -\lambda  }(\lambda e^{ \lambda  }+e^{ \lambda  })\  \ = { \lambda  }^{ 2 }+ \lambda \ \ Variance= E[{ X }^{ 2 }]- { E[X] }^{ 2 }\ = { \lambda  }^{ 2 }+ \lambda - { \lambda  }^{ 2 }\ = \lambda \ \ \ $