Tag: probability distributions

Questions Related to probability distributions

If the mean of Poisson distribution is $\displaystyle \frac{1}{2}$, then the ratio of $P(X=3)$ to $P(X=2)$ is 

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

  1. 1:2

  2. 1:4

  3. 1:6

  4. 1:8

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Correct Option: C
Explanation:

$P(X=3)/P(X=2)$
$=\dfrac{\lambda^{3}(2!)}{3!(\lambda^{2})}$
$=\dfrac{\lambda}{3}$
$=\dfrac{1}{2(3)}$
$=\dfrac{1}{6}$
The ratio is $1:6$

In a poisson distribution, the probability of $0$ success is $10$%. The mean of the distribution is equal to

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

  1. $\log _{10}e$

  2. $\log _{e}10$

  3. $0$

  4. $\dfrac{1}{10}$

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Correct Option: B
Explanation:

Given $P(X=0)=.1$
$\Rightarrow e^{-\lambda}=\cfrac{1}{10}\therefore \lambda = \log _e 10$

The parameter $\lambda $ of poisson distribution is always

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

  1. zero

  2. 1

  3. -1

  4. a finite positive value

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Correct Option: D
Explanation:

In Poisson distribution the  parameter $(  > 0)$, it is always greater than zero or a finite positive value.

If X is a poisson variable with parameter 0.09,then its S.D. is

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

  1. 0.009

  2. 0.3

  3. 0.03

  4. 0.09

Show answer
Correct Option: B
Explanation:

X is a Poisson variable with parameter 0.09.
In Poisson distribution,  Poisson variable(X) is equal to  expected value of X and also to its variance. 
 S.D =  $\sqrt {variance}  =  \sqrt { X } =\sqrt { 0.09 } =0.3$

The standard deviation of P.D. is 1.5, then its mean is

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

  1. 1.5

  2. 2

  3. 2.25

  4. 3.25

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Correct Option: C
Explanation:

Given $\sigma =1.5$
We know that for a Poisson's distribution,mean and variance are equal.
$\mu=\sigma^2$
$\Rightarrow \mu =2.25$

If the mean of poisson distribution is $16$, then its S.D. is

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

  1. $16$

  2. $4$

  3. $10$

  4. $15$

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Correct Option: B
Explanation:

$S.D= \sqrt { \lambda  }$

Mean$= \lambda$
Therefore S.D.$= \sqrt { 16   } \=4 $

Six coins are tossed $6400$ times. The probability of getting $6$ heads $x$ times using poison distribution is

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

  1. $6400{e^{ - x}}$

  2. $\frac{{6400{e^{ - x}}}}{{x!}}$

  3. $\frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$

  4. ${e^{ - 100}}$

Show answer
Correct Option: C
Explanation:

Therefore, the probability of getting $6$ heads with $6$ coins $ = {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{64}} = P\left( { > {a _y}} \right)$
Then, 
$\eta P = 6400 \times \frac{1}{{64}} = 100 = m\left( {{a _y}} \right)$
So, by poison's law, $P\left( {x = n} \right) = \frac{{{e^{ - m}}{m^x}}}{{x!}}$
$ = \frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$

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 } $

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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