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

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:

  1. ${\left( {0.8} \right)^8}$
  2. $3{\left( {0.8} \right)^8}$
  3. $1-{\left( {0.8} \right)^8}$
  4. $1-3{\left( {0.8} \right)^8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

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

  1. $\dfrac { 4 }{ 7 } $
  2. $\dfrac { 3 }{ 7 } $
  3. $\dfrac { 2 }{ 7 } $
  4. $\dfrac { 1 }{ 7 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let x_i be the number of bags with at least i defective bulbs. Total number of defective bulbs = sum(x_i) for i from 1 to 5. Given: 7 bags have >=1, 6 have >=2, 5 have >=3, 4 have >=4, 3 have >=5. Sum of defective bulbs = 7 + 6 + 5 + 4 + 3 = 25. Total bulbs across 20 bags = 20 * 10 = 200. Non-defective bulbs = 200 - 25 = 175. The ratio of defective to non-defective bulbs = 25 / 175 = 1 / 7.

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

For a Poission distribution which pair has same value.

  1. (Mean, Std. Deviation)

  2. (Variance, Standard Deviation)

  3. (Mean, Variance)

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

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

  1. $2$
  2. $3$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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$=\lambda=1$
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

For a Poission distribution, which of the following is true

  1. $Mean = Mode$
  2. $Median = S.D.$
  3. $Mean = Variance$
  4. $Median = Variance$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 
$\therefore \text{mean}=\text{variance}$
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

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

  1. $\displaystyle \frac{6}{5^{e}}$
  2. $\displaystyle \frac{5}{6}$
  3. $\displaystyle \frac{6}{55}$
  4. $\displaystyle \frac{6}{e^{5}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $
$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$

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

The probability of r successes in case of poissons distrbution is

  1. $\dfrac{e^{\gamma }m}{\angle \gamma }$
  2. $\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$
  3. $\dfrac{e^{m}\gamma }{\angle \gamma }$
  4. $\dfrac{e^{-m}m^{r}}{\angle \gamma }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 } $ 

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

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

  1. $2/e^{2}$
  2. $0$
  3. $1-\dfrac{3}{e^{2}}$
  4. $\dfrac{3}{e^{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } } $

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

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

  1. $2\alpha $
  2. $\dfrac{\alpha }{3}$
  3. $\alpha e^{-2}$
  4. $\alpha e^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $  

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

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

  1. $\lambda $
  2. $\sqrt{\lambda }$
  3. $\dfrac{1}{\lambda}$
  4. $\dfrac{1}{\sqrt {\lambda}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 \ \ \ $