Tag: business maths

Questions Related to business maths

A manufactured product on an average has $2$ defects per unit of product produced. If the number of defects follows P.D., the probability of finding zero defects is

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

  1. $e^{-2}$

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

  3. $\displaystyle \frac{e^{-2}2^{1}}{\angle 1}$

  4. $e^{-002}$

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Correct Option: A
Explanation:

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
: 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; ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is 
$ \mu = 2$ (defects per unit of product produced) 
x = 0 (zero defects)
$ P(0; 2)=\dfrac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } = {e}^{-2}$

If the number of telephone calls coming into a telephone exchange between 10 AM and 11 AM follows Poisson distribution with parameter 2 then the probability of obtaining at least one call in that time interval is 

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

  1. $e^{-2}$

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

  3. $2e^{-2}$

  4. $3e^{-2}$

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 $
$ P(x \ge 1; \mu) = 1 - P (x=0 ; \mu) $
$ \mu = 2 $    
$x = 0$ (No call comes) 
$ P(x \ge 1; 2) = 1 - P (x=0 ; 2) = 1 - \dfrac { { e }^{ -2 }{ 2 }^{ 0} }{ 0! } = 1 - {e}^{-2}$
                             

Cycle tyres are supplied in lots of $10$ and there is a chance of $1$ in $500$ to be defective. Using poisson distribution, the approximate number of lots containing no defectives in a consignment of $10,000$ lots if $e^{-0.02}=0.9802$ is

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

  1. $9980$

  2. $9998$

  3. $9802$

  4. $9982$

Show answer
Correct Option: C
Explanation:

Here $\lambda = \cfrac{1}{500}\times 10 = 0.02$
Thus probability that  lot is not defective is $=P(X=0)=\cfrac{e^{-0.002}(.0020^0}{0!}=e^{-.002}=0.9802$
Hence number of no defective lots out of $10,000$ is $=.9802\times 10,000=9802$

The chance of a traffic accident in a day attributed to a taxi driver is $0.001$. Out of a total of $1000$ days the number of days with no accident is

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

  1. $1000\times e^{-1}$

  2. $1000\times e^{-0.1}$

  3. $1000\times e^{-0.001}$

  4. $1000\times e^{-0.0001}$

Show answer
Correct Option: C
Explanation:

Here $\lambda = 0.001$
Hence number of day out of 1000 days without accident is $1000\times P(X=0)=1000\times e^{-0.001}$

A manufacturer of cotter pins knows that $5$% of his product is defective. If he sells cotter pins in boxes of $100$ and guarantees that not more than $10$ pins will be defective, the approximate probability that a box will fail to meet the guaranteed quality is

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

  1. $\displaystyle \frac{e^{-5}5^{10}}{ 10!}$

  2. $1-\displaystyle \sum _{x=0}^{10}\frac{e^{-5}5^{x}}{ x!}$

  3. $1-\displaystyle \sum _{x=0}^{\infty }\frac{e^{-5}5^{x}}{ x!}$

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

Show answer
Correct Option: B
Explanation:

we are given $n=100$
let $p=$ probability of a defective bulb $=5$%$=0.05$
$\therefore m=$ mean number of defective bulbs in a box of $100=np=100\times 0.05=5$
Since p is small , we can use poison's distribution.
Probability of $x$ defective bulbs in a box of $100$ is
$\displaystyle P\left( X=x \right) =\frac { { e }^{ -m }{ m }^{ x } }{ x! } =\frac { { e }^{ -5 }{ 5 }^{ x } }{ x! } ,x=0,1,2...$
Probability that is box will fail to meet the guarented quality is $\displaystyle P\left( X>10 \right) =1-P\left( X\le 10 \right) =1-\sum _{ x=0 }^{ 10 }{ \frac { { e }^{ -5 }{ 5 }^{ x } }{ x! }  } =1-{ e }^{ -5 }\sum _{ x=0 }^{ 10 }{ \frac { { 5 }^{ x } }{ x! }  } $

The number of accidents in a year attributed to a taxi driver in a city follows Poisson distribution with mean $3$. Out of $1000$ taxi drivers, the approximate number of drivers with no accident in a year given that $e^{-3}=0.0498$ is

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

  1. $4.98$

  2. $49.8$

  3. $498$

  4. $4.8$

Show answer
Correct Option: B
Explanation:

Here $\lambda = 3$
Hence Number of drivers with no accident out of 1000 is $=1000\times P(X=0)=1000\times e^{-3}=1000\times 0.0498=49.8$

A manufacturing concern employing a large number of workers finds that, over a period of time, the average absentee rate is $2$ workers per shift. The probability that exactly $2$ workers will be absent in a chosen shift at random is

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

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

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

  3. $e^{-2}$

  4. $e^{-3}$

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 $  = 2 
x = 2 (exact 2 workers)
$ P(2;2)=\dfrac { { e }^{ -2 }{ 2 }^{2 } }{ 2! } $

A manufacturer who produces medicine bottles finds that $0.1$% of the bottles are defective. The bottles are packed in boxes containing $500$ bottles. A drug manufacturer buys $100$ boxes from the producer of bottles. Using poisson distribution,the number of boxes with at least one defective bottle is

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

  1. $100(1-e^{-0.1})$

  2. $100(1-e^{-0.5})$

  3. $100(1-e^{-0.05})$

  4. $100(1-e^{-0.01})$

Show answer
Correct Option: B
Explanation:

Here $\lambda = \cfrac{0.1}{100}\times 500=0.5 $
Hence number of boxes out of 100 which contain at least one defective bottle is,
$=100\left(1-P(X=0)\right)=100\left(1-e^{-0.5}\right)$ 

Suppose $2$% of the people on an average are left handed. The probability of 3 or more left handed among 100 people is

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

  1. $3e^{-2}$

  2. $4e^{-2}$

  3. $1-5e^{-2}$

  4. $5 e^{-2}$

Show answer
Correct Option: C
Explanation:

Here P.D parameter $\lambda = \cfrac{2}{100}\times 100=2$
Hence probability that 3 or more people are left handed is $=1-P(X=0)-P(X=1)-P(X=2)$
$=1-e^{-2}-\cfrac{e^{-2}2}{1!}-\cfrac{e^{-2}2^2}{2!}=1-5e^{-2}$

Suppose there is an average of $2$ suicides per year per $50,000$ population. In a city of population $1,00,000$, the probability that in a given year there are, zero suicides is

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

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

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

  3. $e^{-4}$

  4. $1-e^{-4}$

Show answer
Correct Option: C
Explanation:

P.D parameter $\lambda = \cfrac{2}{50000}\times 100000 = 4$
Thus probability that in a year there is no suicide is $=P(X=0)=e^{-4}$