CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 5 - MCQExams.com

If for a poisson distribution $$P(X=0)=0.2$$, then the variance of the distribution is
  • $$5$$
  • $$log_{10}5$$
  • $$log_{e}5$$
  • $$log_{5}e$$
If $$X$$ is a poisson variate with $$P(X=0)=P(X=1)$$, then $$P(X=2)$$ is
  • $$\dfrac{e}{2}$$
  • $$\dfrac{e}{6}$$
  • $$\dfrac{1}{6e}$$
  • $$\dfrac{1}{2e}$$
A random variable $$X$$ has Poisson distribution with mean $$2$$. Then $$P(X > 1.5)$$ equals
  • $$2/e^{2}$$
  • $$0$$
  • $$1-\dfrac{3}{e^{2}}$$
  • $$\dfrac{3}{e^{2}}$$
If $$X$$ is a random poisson variate such that $$\alpha =p(X=1)=p(X=2)$$, then $$p(X=4)=$$
  • $$\alpha e^{-2}$$
  • $$\alpha e^{2}$$
  • $$2\alpha $$
  • $$\dfrac{\alpha }{3}$$
If $$X$$ is a random poisson variate such that $$E(X^{2})=6$$, then $$E(X)=$$
  • $$-3$$
  • $$2$$
  • $$-3\&2$$
  • $$-2$$
If $$X$$ is a poisson variate with $$P(X=0) = 0.8,$$ then the variance of $$X$$ is
  • $$log_{e}20$$
  • $$log_{10}20$$
  • $$log_{e}(5/4)$$
  • $$0$$
If $$X$$ is a poisson variate such that $$P(X=0)=0.1,P(X=2)=0.2$$, then the parameter $$\lambda $$
  • $$2$$
  • $$4$$
  • $$5$$
  • $$3$$
If $$X$$ is a poisson variate such that $$P(X=0)=P(X=1)$$,then $$P(X=2)=$$
  • $$\dfrac{e}{2}$$
  • $$\dfrac{e}{6}$$
  • $$\dfrac{1}{6e}$$
  • $$\dfrac{1}{2e}$$
If $$X$$ is a poisson variate such that $$P(X=0)=\dfrac{1}{2}$$, the variance of $$X$$ is
  • $$\dfrac{1}{2}$$
  • $$2$$
  • $$\log_{e}2$$
  • $$3$$
If $$X$$ is a Poisson variate such that $$P(X = 2) = 9P(X = 4)$$ then mean and variance of $$X$$ are
  • $$1, 1$$
  • $$2, 2$$
  • $$\displaystyle \frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$$
  • $$\displaystyle\frac{2}{3},\frac{2}{3}$$
If $$X$$ is a Poisson variate such that $$P(X=1) = P(X=2)$$ then $$P(X=4)=$$
  • $$\dfrac{1}{2e^{2}}$$
  • $$\dfrac{1}{3e^{2}}$$
  • $$\dfrac{2}{3e^{2}}$$
  • $$\dfrac{1}{e^{2}}$$
If a random variable $$X$$ follows a P.D. such that $$P(X=1)=P(X=2)$$, then $$P(X=0)=$$
  • $$e^{2}$$
  • $$\dfrac{1}{e^{2}}$$
  • $$\dfrac{1}{e}$$
  • $$e$$
For a Poisson variate $$X$$ if $$P(X=2)=3P(X=3)$$, then the mean of $$X$$ is
  • $$1$$
  • $$1/2$$
  • $$1/3$$
  • $$1/4$$
If for a poisson variable $$ X$$, $$P(X=1)=2.\ P(X=2)$$, then the parameter $$\lambda $$ is
  • $$0$$
  • $$1$$
  • $$2$$
  • $$3$$
In a poisson distribution $$P(X=0)=P(X=1)=k$$, then the value of $$k$$ is
  • $$1$$
  • $$\displaystyle\frac{1}{e}$$
  • $$e$$
  • $$\sqrt{2}$$
If in a poisson distribution $$P(X=1)=P(X=2)$$; the mean of the distribution $$f(x)=e^{-x}\dfrac{\lambda ^{x}}{\angle x}$$ is
  • $$1$$
  • $$2$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{3}{2}$$
If $$X$$ is a Poisson variate with parameter $$\displaystyle \frac{3}{2}$$, find $$P(X\geq 2)$$
  • $$\displaystyle \frac{5}{2}e^{\frac{-3}{2}}$$
  • $$\displaystyle 1-\frac{5}{2}e^{\frac{-3}{2}}$$
  • $$\displaystyle 1-e^{\frac{-3}{2}}$$
  • $$\displaystyle e^{\frac{-3}{2}}$$
If $$X$$ is a Poisson variate with parameter $$1.5$$, then $$P(X>1)$$ is
  • $$1-e^{-1.5}$$
  • $$e^{-1.5}(2.5)$$
  • $$1-e^{-1.5}(2.5)$$
  • $$1-e^{-1.5}(3.5)$$
Suppose $$X$$ is a poisson variable such that $$P(X=2)=\frac{2}{3}P(X=1)$$, then $$P(x=0)$$ is
  • $$\dfrac{3}{4}$$
  • $$e^{\dfrac{4}{3}}$$
  • $$e^{\dfrac{-4}{3}}$$
  • $$\dfrac{1}{2}$$
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
  • $$4$$
  • $$3$$
  • $$1.5$$
  • $$0$$
If $$X$$ is a poisson variable such that $$P(X=2)=\frac{2}{3}P(X=1)$$, then $$P(x=3)$$ is
  • $$e^{\frac{-4}{3}}$$
  • $$\frac{64}{162}e^{\frac{-4}{3}}$$
  • $$e^{\frac{-3}{4}}$$
  • $$e^{\frac{3}{4}}$$
If $$X$$ is a random Poisson variate such that $$P(X=0)=\displaystyle\frac{1}{e}$$, then the variance of the same distribution is
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
In a Poisson distribution, the probability $$P(X=0)$$ is twice the probability $$P(X=1)$$. The mean of the distribution is
  • $$\displaystyle \frac{1}{4}$$
  • $$\displaystyle \frac{1}{3}$$
  • $$\displaystyle \frac{1}{2}$$
  • $$\displaystyle \frac{3}{4}$$
In a binomial distribution $$n = 200, p = 0.04$$. Taking Poisson distribution as an approximation to the binomial distribution .
Assertion (A) :- Mean of the Poisson distribution $$= 8$$
Reason (R) : In a Poisson distribution, $$\displaystyle P(X=4)=\frac{512}{3e^{8}}$$
  • both A and R are true and R is the correct explanation of A
  • both A and R are true and R is not correct explanation of A
  • A is true but R is false
  • A is false but R is true
If $$X$$ is a random poission variate such that $$2P(X=0)+P(X=2)=2P(X=1)$$ then $$E(X)=$$
  • $$4$$
  • $$3$$
  • $$2$$
  • $$1$$
If $$3$$% of electric bulbs manufactured by a company are defective, the probability that a sample of $$100$$ bulbs has no defective bulbs is
  • 0
  • $$e^{-3}$$
  • $$1-e^{-3}$$
  • $$3e^{-3}$$
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
  • $$\displaystyle \frac{e^{-0.02}.2^{1}}{1!}$$
  • $$1-e^{-0.2}$$
  • $$e^{-0.2}$$
  • $$1-e^{1.2}$$
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
  • $$e^{-1.5}$$
  • $$1.5\times e^{-1.5}$$
  • $$1-2.5\times e^{-1.5}$$
  • $$1-1.5\times e^{-1.5}$$
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
  • $$1.12 \times e^{-1.5}$$
  • $$1-2.5 \times e^{-1.5}$$
  • $$1-3.625 \times e^{-1.5}$$
  • $$3.625 \times e^{-1.5}$$
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
  • $$9980$$
  • $$9998$$
  • $$9802$$
  • $$9982$$
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
  • $$\displaystyle e^{-10}\left [ 1+\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$$
  • $$\displaystyle e^{-10}\left [ 1+\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$$
  • $$\displaystyle e^{-10}\left [ 1-\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$$
  • $$\displaystyle e^{-10}\left [ 1-\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$$
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
  • $$e^{-2}$$
  • $$1-e^{-2}$$
  • $$\displaystyle \frac{e^{-2}2^{1}}{\angle 1}$$
  • $$e^{-002}$$
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 
  • $$e^{-2}$$
  • $$1-e^{-2}$$
  • $$\displaystyle \frac{e^{-2}2^{1}}{1!}$$
  • $$e^{-0.02}$$
Suppose $$300$$ misprints are distributed randomly throughout a book of $$500$$ pages. The probability that a given page contains, at least one misprint is 
  • $$1.e^{-0.6}$$
  • $$1-e^{-0.6}$$
  • $$(0.6)e^{-0.6}$$
  • $$(0.06)e^{-0.6}$$
On an average, a submarine on patrol sights $$6$$ enemy ships per hour. Assuming the number of ships sighted in a given length of time is a poisson variate, the probability of sighting atleast one ship in the next $$15$$ minutes is
  • $$e^{-15}$$
  • $$1-e^{-6}$$
  • $$1-e^{-15}$$
  • $$e^{-6}$$
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
  • $$1.e^{-2}$$
  • $$1-e^{-2}$$
  • $$e^{-4}$$
  • $$1-e^{-4}$$
Suppose $$2$$% of the people on an average are left handed. The probability of 3 or more left handed among 100 people is
  • $$3e^{-2}$$
  • $$4e^{-2}$$
  • $$1-5e^{-2}$$
  • $$5 e^{-2}$$
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 no defective bottle is
  • $$100\times e^{-0.1}$$
  • $$100\times e^{-0.5}$$
  • $$100\times e^{-0.05}$$
  • $$100\times e^{-0.01}$$
It is known that the probability that an item produced by a certain machine will be defective is $$0.01$$. Use Poisson distribution, to find the probability in a sample of $$100$$ items selected at random from the total output, that there are not more than one defective item.
  • $$\displaystyle \frac{3}{e}$$
  • $$\displaystyle \frac{2}{e}$$
  • $$\displaystyle \frac{1}{e}$$
  • $$\displaystyle \frac{4}{e}$$
If the number of telephone calls coming into a telephone exchange between 10 AM and 11 AM follows P.D. with parameter 2, then the probability of obtaining zero calls in that time interval is
  • $$e^{-2}$$
  • $$1-e^{-2}$$
  • $$2.e^{-2}$$
  • $$3.e^{-2}$$
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 
  • $$e^{-2}$$
  • $$(1-e^{-2})$$
  • $$2e^{-2}$$
  • $$3e^{-2}$$
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
  • $$\displaystyle \frac{e^{-2}2^{2}}{ 2!}$$
  • $$\displaystyle \frac{e^{-2}2^{3}}{3!}$$
  • $$e^{-2}$$
  • $$e^{-3}$$
Suppose $$220$$ misprints are distributed randomly throughout a book of $$200$$ pages. The probability that a given page contains, no misprint is
  • $$1.e^{-02.2}$$
  • $$1-e^{-02.2}$$
  • $$e^{-1.1}$$
  • $$e^{-2.1}$$
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
  • $$100(1-e^{-0.1})$$
  • $$100(1-e^{-0.5})$$
  • $$100(1-e^{-0.05})$$
  • $$100(1-e^{-0.01})$$
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
  • $$1000\times e^{-1}$$
  • $$1000\times e^{-0.1}$$
  • $$1000\times e^{-0.001}$$
  • $$1000\times e^{-0.0001}$$
On the average a submarine on patrol sights $$6$$ enemy ships per hour. Assuming the number of ships sighted in a given length of time is a poisson variate, the probability of sighting $$4$$ ships in the next two hours is
  • $$\displaystyle \frac{e^{-12}12^{4}}{ 4!}$$
  • $$\displaystyle \frac{e^{-4}12^{12}}{ 3!}$$
  • $$\displaystyle \frac{e^{-6}12^{4}}{ 4!}$$
  • $$\displaystyle \frac{e^{-3}12^{2}}{ 4!}$$
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
  • $$4.98$$
  • $$49.8$$
  • $$498$$
  • $$4.8$$
Four coins are tossed together. What is the probability that head will appear on at least one of the four?
  • $$\displaystyle\frac{15}{16}$$
  • $$\displaystyle\frac{1}{16}$$
  • $$\displaystyle\frac{1}{4}$$
  • None of these
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
  • $$\displaystyle \frac{e^{-5}5^{10}}{ 10!}$$
  • $$1-\displaystyle \sum_{x=0}^{10}\frac{e^{-5}5^{x}}{ x!}$$
  • $$1-\displaystyle \sum_{x=0}^{\infty }\frac{e^{-5}5^{x}}{ x!}$$
  • $$\displaystyle \sum_{x=0}^{\infty }\frac{e^{-5}5^{x}}{ x!}$$
On an average, a submarine on patrol sights $$6$$ enemy ships per hour. Assuming the number of ships sighted in a given length of time is a Poisson variate, the probability of sighting at least two ships in the next $$20$$ minutes is
  • $$1-e^{-2}$$
  • $$1-2e^{-2}$$
  • $$1-3e^{-2}$$
  • $$1-4e^{-2}$$
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers