Tag: business maths

Questions Related to business maths

If, in a Poisson distribution $P(X= 0)=k$ then the variance is: 

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

  1. $e^{\lambda}$

  2. $\log \dfrac{1}{k}$

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

  4. $\log k$

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Correct Option: B

The incidence of an occupational disease to the workers of a factory is found to be $\displaystyle \frac{1}{5000}$ . If there are $10,000$ workers in a factory then the probability that none of them will get the disease is

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

  1. $e^{-1}$

  2. $e^{-2}$

  3. $e^{3}$

  4. $e^{4}$

Show answer
Correct Option: B
Explanation:
By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region.
$x: $ The actual number of successes that occur in a specified region.
$P(x;$$ \mu $ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
$ \mu = 10000 \times \dfrac{1}{5000} = 2 $
$x = 0$
$ P(0;2)=\dfrac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } =  { e }^{ -2 }$

The probability that atmost $5$ defective fuses will be found in a box of $200$ fuses, if experience shows that $20 \%$ of such fuses are defective,  is

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

  1. $\displaystyle \frac{e^{-40}40^{5}}{ 5!}$

  2. $\displaystyle \sum _{x=0}^{5}\frac{e^{-40}40^{x}}{ x!}$

  3. $\displaystyle \sum _{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$

  4. $1-\displaystyle \sum _{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$

Show answer
Correct Option: B
Explanation:

The poisson distribution is
$\displaystyle P(X=x)=\frac{e^{-\lambda }\lambda ^{x}}{x!}$ , $x=0,1,2,3,..$
Let, $X$ denote the defective fuse

$p=\frac{20}{100}$
$n=200$
mean$ = \lambda  = np = 200\times \frac{20}{100} = 40$
$\displaystyle => P($atmost $5$ defective fuses)$= _{x=0}^{5}\sum \frac{e^{-40}40^{x}}{x!}$

There are $500$ boxes each containing $1000$ ballot papers for election. The chance that a ballot paper is defective is $0.002$. Assuming that the number of defective ballot papers follow Poisson distribution, the number of boxes containing at least one defective ballot paper given that $e^{-2}=0.1353$ is

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

  1. $216$

  2. $432$

  3. $648$

  4. $234$

Show answer
Correct Option: B
Explanation:

Here $\lambda = .002\times 1000=2$
Thus probability that a box contain at least one defective ballot $=1-P(X=0)=1-e^{-2}=1-0.1353=.8647$   
Hence   number of box out of $500$ which contain at least one defective ballot is $=.8647\times 500 =432$

Six unbiased coins are tossed $6400$ times. Using Poisson distribution, the approximate probability of getting six heads $2$ times is

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

  1. $\displaystyle \frac{e^{-64}(64)^{2}}{ 2!}$

  2. $\displaystyle \frac{e^{-100}(100)^{2}}{ 2!}$

  3. $1-\displaystyle \frac{e^{-100}(100)^{x}}{ x!}$

  4. $\displaystyle \frac{e^{-100}(100)^{x}}{x!}$

Show answer
Correct Option: B
Explanation:

Here $\lambda = \left(\cfrac{1}{2}\right)^6\times 64100=100$
Hence probability of getting six heads two times $=P(X=2)=\cfrac{e^{-100}(100)^2}{2!}$

A company knows on the basis of past experience that $2$% of the blades are defective. The probability of having 3 defective blades in a sample of $100$ blades is

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

  1. $e^{-2}2^{2}$

  2. $\displaystyle \frac{e^{-2}2^{3}}{3!}$

  3. $\displaystyle \frac{e^{-2}2^{3}}{ 2!}$

  4. $\displaystyle \frac{e^{-4}2^{-1}}{ 2!}$

Show answer
Correct Option: B
Explanation:

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
Here,$ \mu  =  100 \times 0.02 = 2 $
$x = 3$(defective blades)
$ P(3;2)=\dfrac { { e }^{ -2 }{ 2 }^{3 } }{ 3! } $

A car hire firm has $2$ cars which it hires out day by day. If the number of demands for a car on each day follows poisson distribution with parameter $1.5$, then the probability that neither car is used is

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

  1. $e^{-1.5}$

  2. $1.5\times e^{-1.5}$

  3. $1-2.5\times e^{-1.5}$

  4. $1-1.5\times e^{-1.5}$

Show answer
Correct Option: A
Explanation:

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
Here, $ \mu  = 1.5$ and $x = 0$ (neither car is used)
$ P(0;1.5)=\dfrac { { e }^{ -1.5 }{ 1.5 }^{0 } }{ 0! } = {e}^{-1.5} $

In a big city, $5$ accidents take place over a period of $100$ days. If the numebr of accidents follows P.D., the probability that there will be $2$ accidents in a day is

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

  1. $\displaystyle \frac{e^{-5}5^{2}}{ 2!}$

  2. $\displaystyle \frac{e^{-05}5^{2}}{ 2!}$

  3. $\displaystyle \frac{e^{-005}(0.05)^{2}}{ 2!}$

  4. $\displaystyle \frac{e^{5}5^{2}}{ 2!}$

Show answer
Correct Option: C
Explanation:
By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$\mu\ :$ The mean number of successes that occur in a specified region.
$x\ :$ The actual number of successes that occur in a specified region.
$P(x;\mu )\ :$ The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is 
$ \mu = \dfrac {5}{100} = 0.05 $
$x = 2$ 
$ P(2; 005)=\dfrac { { e }^{ -0.05 }{ 0.05 }^{ 2 } }{ 2! } $

If ${ \mu  } _{ 2 }=20,{ \mu  } _{ 2 }^{ 1 }=276$ for a discrete random variable $X$, then the mean of the random variable $X$ is

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

  1. $16$

  2. $5$

  3. $2$

  4. $1$

Show answer
Correct Option: A
Explanation:

$\mu _2=E(X^2)-[E(X)]^2......(i)$

$\mu _2'=E(X)^2.....(ii)$
Given: $\mu _2=20, \mu _2'=276$
from equation (ii) $\implies E(x^2)=276$
and from equation (i), $\implies 20=276-[E(X)]^2$
$\implies [E(X)]^2=276-20=250\\implies E(X)=16$
is the mean of random variable.