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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
  • log105
  • loge5
  • log5e
If X is a poisson variate with P(X=0)=P(X=1), then P(X=2) is
  • e2
  • e6
  • 16e
  • 12e
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


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