Tag: random variables and probability distribution

Quizzes (1) Questions (91)

Questions Related to random variables and probability distribution

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

If the probability of that a poisson variable $X$ takes a positive value $\geq 1$ is $1-e^{-1.5}$, then the varianceof the distribution is

  1. $4$
  2. $3$
  3. $1.5$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $P(X\geq 1) = 1-e^{-1.5}$
$\Rightarrow 1-P(X=0)=1-e^{-1.5}\Rightarrow P(X=0)=e^{-1.5}=e^{-\lambda}\therefore \lambda = 1.5$

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

In a town $10$ accidents take place in a span of $50$ days. Assuming that number of accidents follows Poisson distribution, the probability that there will be atleast one accident on a selected day at random is

  1. $\displaystyle \frac{e^{-0.02}.2^{1}}{1!}$
  2. $1-e^{-0.2}$
  3. $e^{-0.2}$
  4. $1-e^{1.2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here Poisson parameter $\lambda = \cfrac{10}{50}=0.2$
$\therefore P(x \geq 1)=1-P(X=0)=1-\cfrac{e^{-0.2}.(.2)^0}{0!}=1-e^{-0.2}$ 

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

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 both the cars is used is

  1. $1.12 \times e^{-1.5}$
  2. $1-2.5 \times e^{-1.5}$
  3. $1-3.625 \times e^{-1.5}$
  4. $3.625 \times e^{-1.5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = 1$
Hence probability that both the cars are used $=1-P(X=0)-P(X=1)=1-\cfrac{e^{-1.5}(1.5)^0}{0!}-\cfrac{e^{-1.5}(1.5)^1}{1!}=1-2.5e^{-1.5}$

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

If $X$ is a Poisson variate with parameter $\displaystyle \frac{3}{2}$, find $P(X\geq 2)$

  1. $\displaystyle \frac{5}{2}e^{\frac{-3}{2}}$
  2. $\displaystyle 1-\frac{5}{2}e^{\frac{-3}{2}}$
  3. $\displaystyle 1-e^{\frac{-3}{2}}$
  4. $\displaystyle e^{\frac{-3}{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = \cfrac{3}{2}$
$\therefore P(X\geq 2) = 1-P(X=0)-P(X=1)$
$\displaystyle =1-\cfrac{e^{-3/2}(3/2)^0}{0!}-\cfrac{e^{-3/2}(3/2)^1}{1!}=1-\cfrac{5}{2}e^{-3/2}$

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

If $X$ is a random Poisson variate such that $P(X=0)=\displaystyle\frac{1}{e}$, then the variance of the same distribution is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=\frac{1}{e}$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=\dfrac{1}{e}$
$\lambda=1$
Hence mean=variance=$1$

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

If on an average ,5 percent of the output in a factory making certain parts, is defective and that 200 units are in a package then the probability that atmost 4 defective parts may be found in that package is

  1. $\displaystyle e^{-10}\left [ 1+\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$
  2. $\displaystyle e^{-10}\left [ 1+\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$
  3. $\displaystyle e^{-10}\left [ 1-\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$
  4. $\displaystyle e^{-10}\left [ 1-\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda$
$=\dfrac{5}{100}.200$
$=10$.
Hence by applying Poisson distribution, we get that the probability that atmost 4 defective part are found is 
$=\sum _{k=0} ^{k=4} \dfrac{e^{-10}.10^{k}}{k!}$

$=e^{-10}[1+\dfrac{10}{1!}+\dfrac{10^{2}}{2!}+\dfrac{10^{3}}{3!}+\dfrac{10^{4}}{4!}]$.

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

Suppose $300$ misprints are distributed randomly throughout a book of $500$ pages. The probability that a given page contains, at least one misprint is 

  1. $1.e^{-0.6}$
  2. $1-e^{-0.6}$
  3. $(0.6)e^{-0.6}$
  4. $(0.06)e^{-0.6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = \cfrac{300}{500}=0.6$
Hence $ P(X\geq 1) = 1-P(X=0)=1-\cfrac{e^{-0.6}(0.6)^0}{0!}=1-e^{-0.6}$

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

A manufactured product on an average has 2 defects per unit of product produced. If the number of defects follows Poisson distribution, the probability of finding at least one defect is 

  1. $e^{-2}$
  2. $1-e^{-2}$
  3. $\displaystyle \frac{e^{-2}2^{1}}{1!}$
  4. $e^{-0.02}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $
$ P(x \ge 1; \mu) = 1 - P (x=0 ; \mu) $
$ \mu = 2 $    
$x = 0$ (No defective product) 
$ P(x \ge 1; 2) = 1 - P (x=0 ; 2) = 1 - \dfrac { { e }^{ -2 }{ 2 }^{ 0} }{ 0! } = 1 - {e}^{-2}$
                             

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

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 only one car is used is

  1. $e^{-1.5}$
  2. $1.5\times e^{-1.5}$
  3. $1-2.5\times e^{-1.5}$
  4. $1-1.5\times e^{-1.5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = 1.5$
Hence probability that only one car is used is $=P(X=1) = \cfrac{e^{-1.5}(1.5)}{1!}=1.5e^{-1.5}$

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

If $3$% of electric bulbs manufactured by a company are defective, the probability that a sample of $100$ bulbs has no defective bulbs is

  1. 0
  2. $e^{-3}$
  3. $1-e^{-3}$
  4. $3e^{-3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 = 100 \times 0.03 = 3 $    
$x = 0$ (No defective bulbs) 
$ P(0; 3)=\dfrac { { e }^{ -3 }{ 3 }^{ 0 } }{ 0! }  $
$P(0;3)={ e }^{ -3 } $