MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 8

$X$ follows a binomial distribution with parameters

$n=8$ and

$p=\cfrac{1}{2}$ , then

$P\left( \left| x-4 \right| \le 2 \right)$ is equal to

0%
$\cfrac { 119 }{ 128 }$
0%
$\cfrac { 116 }{ 128 }$
0%
$\cfrac { 29 }{ 128 }$
0%
None of these

Explanation

We have

$\quad P\left( \left| x-4 \right| \le 2 \right) =P\left( -2\le x-4\le 2 \right) =P\left( 2\le x\le 6 \right)$

$=1-\left[ P(x=0)+P(x=1)+P(x=7)+P(x=8) \right]$

$=1-\left[ { _{ }^{ 8 }{ C } }_{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 }+{ _{ }^{ 8 }{ C } }_{ 1 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 }+{ _{ }^{ 8 }{ C } }_{ 7 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 }+{ _{ }^{ 8 }{ C } }_{ 8 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 } \right]$

$=1-18{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 }=\cfrac { 119 }{ 128 }$

Out of

$100$ bicycles,

$10$ bicycles have puncture. What is the probability of not having any punctured bicycle in a sample of

$5$ bicycles?

0%
$\dfrac { 1 }{ { 2 }^{ 9 } }$
0%
${ \left( \dfrac { 9 }{ 10 } \right) }^{ 5 }$
0%
$\dfrac { 1 }{ { 10 }^{ 5 } }$
0%
$\dfrac { 1 }{ { 2 }^{ 5 } }$

Explanation

Here,

$n=5$ ,

$r=0$ ,

$p=\dfrac { 10 }{ 100 }$ ,

$q=\dfrac { 90 }{ 100 }$

Now,

$P\left( X=0 \right) =_{ }^{ n }{ { C }_{ r } }{ p }^{ r }{ q }^{ n-r }$

$=_{ }^{ 5 }{ { C }_{ 0 } }{ \left( \dfrac { 10 }{ 100 } \right) }^{ 0 }{ \left( \dfrac { 90 }{ 100 } \right) }^{ 5-0 }$

$=1\times 1\times { \left( \dfrac { 9 }{ 10 } \right) }^{ 5 }={ \left( \dfrac { 9 }{ 10 } \right) }^{ 5 }$ .

In a trial, the probability of success is twice the probability of failure. In six trial, the probability of atleast four successes will be

0%
$\dfrac {496}{729}$
0%
$\dfrac {400}{729}$
0%
$\dfrac {500}{729}$
0%
$\dfrac {600}{729}$

Explanation

Let the probability of success and failure be

$p$ and

$q$ , respectively.
Then,

$p = 2q$
and

$p + q = 1$

$\Rightarrow 3q =1$

$\Rightarrow q = \dfrac {1}{3}$
Therefore,

$p = 1-\dfrac {1}{3}=\dfrac {2}{3}$
Required probability

$= \ ^{6}C_{4}\left (\dfrac {2}{3}\right )^{4} \left (\dfrac {1}{3}\right )^{2} + \ ^{6}C_{5}\left (\dfrac {2}{3}\right )^{6} \left (\dfrac {1}{3}\right ) + \ ^{6}C_{6} \left (\dfrac {2}{3}\right )^{6}$

$= \dfrac {496}{729}$ .

If the mean and SD of a binomial distribution are

$20$ and

$4$ respectively, then the number of trials is?

0%
$50$
0%
$25$
0%
$100$
0%
$80$

Explanation

We have,

$np=20$ and

$\sqrt{npq}=4\Rightarrow npq=16$

$\Rightarrow q=\displaystyle\frac{4}{5}$

$\Rightarrow p=1-q=\displaystyle\frac{1}{5}$

$\because np=20$

$\therefore n=100$ .

Two dice are tossed

$6$ times, then the probability that

$7$ will show an exactly four of tosses is

0%
$\dfrac {225}{18442}$
0%
$\dfrac {116}{20003}$
0%
$\dfrac {125}{15552}$
0%
None of these

Explanation

Two dice will show

$7$ if they show :

$\{1,6\},\{2,5\},\{6,1\}\{5,2\},\{4,3\},\{3,4\}$

the probability of success

$(p)= 6/(6\times6)=1/6$

the probability of failure

$(q)= 1-1/6=5/6$

Probability of showing 7 exactly 4 times

$=\,^6C_4\times p^4\times q^2\\=15\times (1/6)^4\times (5/6)^2\\=\dfrac{125}{15552}$

The length of similar components produced by a company is approximated by a normal distribution model with a mean of

$5$ cm and a standard deviation of

$0.02$ cm. If a component is chosen at random .what is the probability that the length of this component is between

$4.98$ and

$5.02$ cm?

0%
$0.5826$
0%
$0.6826$
0%
$0.6259$
0%
$0.6598$

Explanation

Given,

$\mu= 5$ (mean)

$\sigma =0.02$ (standard deviation)

find the probability for

$4.98<x<5.02$

converting the problem in standard form

$Z=(x-\mu)/\sigma$

for

$x=4.98$ ,

$Z=-1$

for

$x=5.02$ ,

$Z=1$

for finding the probability for

$4.98<x<5.02$

in the standard form

$-1<z<1$

in standard form mean is equal to zero and the standard deviation is 1.

ao we will have find the probability for

$(\mu - \sigma)$ to

$(\mu+\sigma)$

i.e.

$-1\ to +1$

and probability for

$(\mu - \sigma)$ to

$(\mu + \sigma)$ is

$0.6826$

$X$ is a Normally distributed variable with mean

$= 30$ and standard deviation

$= 4$ . Find

$P(30 < x<35)$

0%
$0.3698$
0%
$0.3956$
0%
$0.2134$
0%
$0.3944$

Explanation

Here we need to determine the value of

$Z=\dfrac{x-\mu}{\sigma}$

Here,

$\mu=\text{Mean}=30$ and

$\sigma=\text{Stand Deviation}=4$

For,

$x=30\Rightarrow Z=\dfrac{30-30}{4}=0$

For,

$x=35\Rightarrow Z=\dfrac{35-30}{4}=1.25$

$\Rightarrow P(30<x<35)=P(0<Z<1.25)$

$\Rightarrow P(0<Z<1.25)=P(Z<1.25)-P(Z<0)$

$(\because P(a<Z<b)=P(Z<b)-P(Z<a))$

From the Normal Distribution table,

$P(Z<1.25)=0.8944$ and

$P(Z<0)=0.5$

$\Rightarrow P(30<x<35)=0.8944-0.5=0.3944$

$X$ is a Normally distributed variable with mean

$= 30$ and standard deviation

$= 4$ . Find

$P(x>21$ )

0%
$0.9878$
0%
$0.9383$
0%
$0.9975$
0%
$0.9126$

Explanation

$P(x>21)=P(Z>(21-30)/4)=P(Z>-2.25)=0.9878$

The value

$0.9878$ is taken from table.

$X$ is a binomial variate with the range

$\left \{0, 1, 2, 3, 4, 5, 6\right \}$ and

$P(X = 2) = 4P(X = 4)$ , then the parameter

$p$ of

$X$ is

0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {2}{3}$
0%
$\dfrac {3}{4}$

Explanation

Here,

$n = 6$

Given,

$P(X = 2) = 4P(X = 4)$

$\therefore ^{6}C_{2}p^{2}q^{4} = 4^{6}C_{4}p^{4}q^{2}$

$\Rightarrow q^{2} = 4p^{2}$

$\Rightarrow (1 - p)^{2} = 4p^{2}$

$\Rightarrow 1^{2} + p^{2} - 2p = 4p^{2}$

$\Rightarrow 3p^{2} + 2p - 1 = 0$

$\Rightarrow (p + 1)(3p - 1) = 0\Rightarrow p = \dfrac {1}{3} - 1$

$\therefore p = \dfrac {1}{3} (\because$ probability cannot be negative).

Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of

$500$ and a standard deviation of

$100$ . Tom wants to be admitted to this university and he knows that he must score better than at least

$70$ % of the students who took the test. Tom takes the test and scores

$585$ . Tom does better than what percentage of students?

0%
$89.23$ %
0%
$77.26$ %
0%
$70.23$ %
0%
$80.23$ %

Explanation

Given,

$\mu= 500$ (mean)

$\sigma =100$ (standard deviation)

Tom does better than what percentage?

Tom got

$585$ marks

we will have to find how many students got less than

$585$

$x<585$ (find the probability)

converting the problem in standard form

$Z=(x-\mu)/\sigma$

for

$x=585$

$Z=(585-500)/100 = 0.85$

$P(Z<0.85)=P(Z<0)+P(0<Z<0.85)$

$P(Z<0)=0.5$

$P(0<Z<0.85)=0.3023$ (from Z- table)

so the total probability

$= 0.5+0.3023 = 0.8023$

so Tom got more than

$80.23\%$ of students

For a certain type of computers, the length of time between charges of the battery is normally distributed with a mean of

$50$ hours and a standard deviation of

$15$ hours. John owns one of these computers and wants to know the probability that the length of time will be between

$50$ and

$70$ hours.

0%
$0.4082$
0%
$0.4025$
0%
$0.4213$
0%
$0.4156$

Explanation

Given,

$\mu= 50$ (mean)

$\sigma =15$ (standard\ deviation)

find the probability for

$50<x<70$

converting the problem in standard form

$Z=\dfrac{(x-\mu)}{\sigma}$

for

$x=50$ ,

$Z=0$

for

$x=70$ ,

$Z=(70-50)/15 = 1.33$

for finding the probability for

$50<x<70$

In the standard form

$0<z<1.33$

using

$Z$ -table, the area is equal to

$0.4082$

In a binomial distribution, the mean is

$15$ and variance of

$10$ . Then, parameter

$n$ is

0%
$28$
0%
$16$
0%
$45$
0%
$25$

Explanation

Given, mean,

$np = 15$
and variance,

$npq = 10$

$\Rightarrow np (1 - p) = 10$

$\Rightarrow 15(1 - p) = 10$

$\Rightarrow 1 - p = \dfrac {2}{3}$

$\Rightarrow p = 1 - \dfrac {2}{3} \Rightarrow p = \dfrac {1}{3}$

$\therefore n\times \left (\dfrac {1}{3}\right ) = 15\Rightarrow n = 45$ .

$X$ is a Normally distributed variable with mean

$= 30$ and standard deviation

$= 4$ . Find

$P(x<40)$

0%
$0.9789$
0%
$0.9938$
0%
$0.9838$
0%
$0.9538$

Explanation

$P(x<40)=P(Z<(40-30)/4)=P(Z<2.5)=0.9938$

The value

$0.9938$ is taken from normal distribution table.

$X$ and

$Y$ are independent binomial variates

$B\left( 5,\cfrac { 1 }{ 2 } \right)$ and

$B\left( 7,\cfrac { 1 }{ 2 } \right)$ , then

$P\left( X+Y=3 \right)$ is

0%
$\cfrac { 35 }{ 47 }$
0%
$\cfrac { 55 }{ 1024 }$
0%
$\cfrac { 220 }{ 512 }$
0%
$\cfrac { 11 }{ 204 }$

Explanation

$B\left( 5,\cfrac { 1 }{ 2 } \right) \Rightarrow n=5,p=\cfrac { 1 }{ 2 } ,q=\cfrac { 1 }{ 2 }$

$B\left( 7,\cfrac { 1 }{ 2 } \right) \Rightarrow n=7,p=\cfrac { 1 }{ 2 } ,q=\cfrac { 1 }{ 2 }$
since

$x$ and

$y$ are independent events

$x+y=3$ means

$x=0,y=3,x=1,y=2;x=2,y=1;x=0,y=3$

$\therefore P(x+y=3)={ _{ }^{ 5 }{ C } }_{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }.{ _{ }^{ 7 }{ C } }_{ 3 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 7 }+{ _{ }^{ 5 }{ C } }_{ 1 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }{ _{ }^{ 7 }{ C } }_{ 2 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 7 }+{ _{ }^{ 5 }{ C } }_{ 2 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }{ _{ }^{ 7 }{ C } }_{ 1 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 7 }+{ _{ }^{ 5 }{ C } }_{ 3 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }{ _{ }^{ 7 }{ C } }_{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 7 }$

$=\cfrac { 55 }{ 1024 }$ (on simplification)

The length of life of an instrument produced by a machine has a normal distribution with a mean of

$12$ months and standard deviation of

$2$ months. Find the probability that an instrument produced by this machine will last less than

$7$ months.

0%
$0.2316$
0%
$0.0062$
0%
$0.0072$
0%
$0.2136$

Explanation

Life of intsrument follows normal distribution.

Given mean i.e.

$\mu$

$= 12$ months

Given standard deviation i.e.

$\sigma$

$= 2$ months

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$z = (X - μ) / σ$

where

$X$ is a normal random variable,

$μ$ is the mean of

$X$ , and

$σ$ is the standard deviation of

$X$ .

For

$X =22$

$Z= (7-12)/2 = -2.5$

$P( X < 7) = P(Z < -2.5)$

$= 0.0062$

Four cards are drawn from a deck of

$52$ cards, the probability of all being spade is.....

0%
$\dfrac{1}{256}$
0%
$\dfrac{1}{56}$
0%
$\dfrac{1}{64}$
0%
$\dfrac{31}{256}$

Explanation

Let

$X$ : be the number of spade cards.

Drawing is a Bernoulli trial.

So,

$X$ has the binomial distribution,

$P(X=x)=^nC_xq^{n-x}p^x$

where n = number of cards drawn = 4, p = probability of getting a spade card =

$\dfrac {13}{52}=\dfrac 14$

Hence,

$q=1-p=1-\dfrac 14=\dfrac 34$

Hence

$P(X=x)=^4C_x\left(\dfrac 34\right)^{4-x}\left(\dfrac 14\right)^x$

Hence the probability of all being spade is

$P(X=4)=^4C_4\left(\dfrac 34\right)^{4-4}\left(\dfrac 14\right)^4=\left(\dfrac 14\right)^4=\dfrac {1}{256}$

A large group of students took a test in Physics and the final grades have a mean of

$70$ and a standard deviation of

$10$ . If we can approximate the distribution of these grades by a normal distribution, what percent of the students should fail the test (grades

$<60$ )?

0%
$15.21$
0%
$23.21$
0%
$15.87$ %
0%
$16.23$

Explanation

Final grades follows normal distribution.

Given mean i.e.

$\mu$

$= 70$

Given standard deviation i.e.

$\sigma$

$= 10$

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$z = (X - μ) / σ$

where

$X$ is a normal random variable,

$μ$ is the mean of

$X$ , and

$σ$ is the standard deviation of

$X$ .

For

$X =60$

$Z= (60-70)/10 = -1$

$P(X < 60) = P(Z < -1)$

$= 0.1587$

Hence percent of students failed in test is

$15.87\%$

The time taken to assemble a car in a certain plant is a random variable having a normal distribution of

$20$ hours and a standard deviation of

$2$ hours. What is the probability that a car can be assembled at this plant in a period of time between

$20$ and

$22$ hours?

0%
$0.3513$
0%
$0.3216$
0%
$0.3413$
0%
$0.3613$

Explanation

TIme taken to assemble follows normal distribution.

Given mean i.e.

$\mu$

$= 20$ hours

Given standard deviation i.e.

$\sigma$

$= 2$ hours

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$z = (X - μ) / σ$

where

$X$ is a normal random variable,

$μ$ is the mean of

$X$ , and

$σ$ is the standard deviation of

$X$ .

For

$X =22$

$Z= (22-20)/2 = 1$

For

$X = 20$

$Z= (20-20)/2 = 0$

$P(a < X < b) = P(X < b) – P( X < a)$ as understood from diagram

$P(20<X<22) = P(X<22) - P(X<20)$

$= P(Z<1) - P(Z<0)$

$= 0.3413$

$X=x$	$0$	1	2	3	4	5	6	7
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$K^2$	$2k^2$	$7k^2+k$

then $P(0<X<5)=$

0%
$\frac{1}{10}$
0%
$\frac{3}{10}$
0%
$\frac{8}{10}$
0%
$\frac{7}{10}$

Explanation

$\sum P(X=x_1)=1$

$9K+10k^2=1$

$10K^2+9K-1=0$

$10K^2+10K-K-1=0$

$10K(k+1)-1(k+1)=0$

$K=\frac{1}{10}$

$P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=8K=\frac{8}{10}$

$\dfrac{^{11}C_0}{1}$ +

$\dfrac{^{11}C_1}{2}$ +

$\dfrac{^{11}C_2}{3}$ + .................... +

$\dfrac{^{11}C_10}{11}$ =

0%
$\dfrac{2^{11} - 1 }{11}$
0%
$\dfrac{2^{11} - 1 }{6}$
0%
$\dfrac{4^{11} - 1 }{11}$
0%
$\dfrac{3^{11} - 1 }{11}$

Explanation

We can use binomial expansion to do this

$(1+x)^{11}=^{11}C_0+^{11}C_{1}x+^{11}C_2x^2.........^{11}C_{10}x^{10}+^{11}C_{11}x^{11}$

Now integrate the following code

$\int_0^1(1+x)^{11}=^{11}C_0+^{11}C_{1}x+^{11}C_2x^2.........^{11}C_{10}x^{10}+^{11}C_{11}x^{11}$

$\rightarrow\dfrac{(1+x)^{12}}{12}-\dfrac{1}{12}=\dfrac{^{11}C_0}{1}+\dfrac{^{11}C_{1}x}{2}+\dfrac{^{11}C_2x^3}{3}.........\dfrac{^{11}C_{10}x^{10}}{11}+\dfrac{^{11}C_{11}x^{12}}{12}$

Now put

$x=1$

$\rightarrow\dfrac{(1+1)^{12}-1}{12}=\dfrac{^{11}C_0}{1}+\dfrac{^{11}C_{1}}{2}+\dfrac{^{11}C_2}{3}.........\dfrac{^{11}C_{10}}{11}+\dfrac{^{11}C_{11}}{12}$

$\rightarrow\dfrac{(2)^{12}-1}{12}-\dfrac{1}{12}=\dfrac{^{11}C_0}{1}+\dfrac{^{11}C_{1}}{2}+\dfrac{^{11}C_2}{3}.........\dfrac{^{11}C_{10}}{11}$

Thus

$\dfrac{^{11}C_0}{1}+\dfrac{^{11}C_{1}}{2}+\dfrac{^{11}C_2}{3}.........\dfrac{^{11}C_{10}}{11}=\dfrac{2^{12}-2}{12}=\dfrac{2^{11}-1}{6}$

The probability that Dhoni will hit century in every ODI matches he plays is

$\dfrac{1}{5}$ .If he plays

$6$ matches in World Cup

$2011,$ the probability that he will score

$2$ centuries is:

0%
$\dfrac {768}{3125}$
0%
$\dfrac {2357}{3125}$
0%
$\dfrac {2178}{3125}$
0%
$\dfrac {412}{3125}$

Explanation

Given: Probability of century :

$\dfrac 15=0.2$ , total

$6$ matches played

To find: the probability that he will score

$2$ centuries

Using binomial distribution :

$n=6, k=2$

$P =^6C_2\left(\dfrac 15\right)^2\left(\dfrac 45\right)^4\\=15\times \dfrac {256}{15625}=\dfrac {768}{3125}$

is the required probability.

A pair of die is rolled up. If the sum of two dies is

$10$ , find the probability that one of the die showed

$4$ .

0%
$\dfrac{1}{3}$
0%
$\dfrac{7}{128}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{7}{28}$

Explanation

Total no. of outcomes when two dice are thrown are:-

$\left( 1,1 \right) \left( 1,2 \right) \left( 1,3 \right) \left( 1,4 \right) \left( 1,5 \right) \left( 1,6 \right)$

$\left( 2,1 \right) \left( 2,2 \right) \left( 2,3 \right) \left( 2,4 \right) \left( 2,5 \right) \left( 2,6 \right)$

$\left( 3,1 \right) \left( 3,2 \right) \left( 3,3 \right) \left( 3,4 \right) \left( 3,5 \right) \left( 3,6 \right)$

$\left( 4,1 \right) \left( 4,2 \right) \left( 4,3 \right) \left( 4,4 \right) \left( 4,5 \right) \left( 4,6 \right)$

$\left( 5,1 \right) \left( 5,2 \right) \left( 5,3 \right) \left( 5,4 \right) \left( 5,5 \right) \left( 5,6 \right)$

$\left( 6,1 \right) \left( 6,2 \right) \left( 6,3 \right) \left( 6,4 \right) \left( 6,5 \right) \left( 6,6 \right)$

Sum of $$10$ in pair of die can be:

$\{(4, 6), (5, 5), (6, 4)\}$

Outcome of sum of

$10$ is

$n(S)=3$

Out of this those showing

$4$ are

$\{(4, 6), (6, 4)\}$ ,

$n(E)=2$

$\therefore$ Probability that one of the die showed

$4$ .

$=\cfrac { 2 }{ 3 }$

Four cards are drawn from a deck of

$52$ cards, the probability of none being spade is........

0%
$\dfrac{1}{256}$
0%
$\dfrac{7}{256}$
0%
$\dfrac{81}{256}$
0%
$\dfrac{31}{256}$

Explanation

Let

$X$ : be the number of spade cards.

Drawing is a Bernoulli trial.

So,

$X$ has the binomial distribution,

$P(X=x)=^nC_xq^{n-x}p^x$

where n = number of cards drawn = 4, p = probability of getting a spade card =

$\dfrac {13}{52}=\dfrac 14$

Hence,

$q=1-p=1-\dfrac 14=\dfrac 34$

Hence

$P(X=x)=^4C_x\left(\dfrac 34\right)^{4-x}\left(\dfrac 14\right)^x$

Hence the probability of none being spade is

$P(X=0)=^4C_0\left(\dfrac 34\right)^{4-0}\left(\dfrac 14\right)^0=\left(\dfrac 34\right)^4=\dfrac {81}{256}$

The probability that Sania wins Wimbledon tournaments final is

$\dfrac{1}{3}$ . If Sania Mirza plays

$3$ round of Wimbledon final, the probability that she losses all the rounds is:

0%
$\dfrac{8}{27}$
0%
$\dfrac{19}{27}$
0%
$\dfrac{10}{27}$
0%
$\dfrac{17}{27}$

Explanation

Probability that Sania wins Wimbledon tournament final

$=\dfrac{1}{3}$

Probability that Sania losses

$=\dfrac{2}{3}$

i.e,

$p=\dfrac{1}{3}$

$q=\dfrac{2}{3}$

Total no. of rounds,

$n=3$

P (She looses in all round ) , By Bernoullsi's theorm

$\Rightarrow P={^nC_r}{p^rq^{n-r}}={^3C_0}\left( \dfrac { 1 }{ 3 } \right) ^{ 0 }\left( \dfrac { 2 }{ 3 } \right) ^{ 3 }$

$=\dfrac{8}{27}.$

Hence, the answer is

$\dfrac{8}{27}.$

If variance of ten observations 10,20,30,40,50.........100 is A and variance of other ten observations 22,42,62,82,102........,202 is B, then

$\dfrac{B}{A}$ is

0%
0
0%
1
0%
2
0%
4

Explanation

10,20,30...100

Mean=

$=\frac { 1 }{ 10 } \sum { X }$ =55

Variance=

$=\frac { 1 }{ 10 } \sum { { (X-m) }^{ 2 } }$

On solving we get variance=30.27=A

22,42,62..202

Mean=

$=\frac { 1 }{ 10 } \sum { X }$ =112

Variance=

$=\frac { 1 }{ 10 } \sum { { (X-m) }^{ 2 } }$

On solving we get variance=60.55=B

Therefore, B/A=2

The number of terms in the expansion of

$(x+ y+z)^{10}$ is

0%
11
0%
33
0%
66
0%
None of these

Suppose x is a binomial distribution with parameters n = 100 and p = 1/2 then P(X=r) is maximum when r =

0%
50
0%
32
0%
33
0%
67

Explanation

${ 1/2 }^{ n}$

In binomial distribution system

n=100,p=1/2

P(X=r)=

$^{ 100 }{ C_{ r } }$ *

${ 1/2 }^{ r }*{ 1/2 }^{ n-r }$

$^{ 100 }{ C_{ r } }$ *

${ (1/2 )}^{ n}$

$^{ n }{ C_{ r } }$ is maximum at r=n/2

Therefore,

$^{ 100 }{ C_{ r } }$ is maximum at r=100/2=50

P(X=r) is maximum when r = 50

A dice is tossed

$5$ times. Getting an odd number is considered a success. Then the variance of distribution of success is

0%
$\dfrac { 8 }{ 3 }$
0%
$\dfrac { 3 }{ 8 }$
0%
$\dfrac { 4 }{ 5 }$
0%
$\dfrac { 5 }{ 4 }$

Explanation

$n=5$

possible outcomes when a dice is tossed once

$={1,2,3,4,5,6}$

Getting an odd number is considered as success,

$p=\dfrac{3}{6}$

$=\dfrac{1}{2}$

Getting even is considered as failure,

$q=1-\dfrac{1}{2}$

$=\dfrac{1}{2}$

Variance

$=npq$

$=5\times\dfrac{1}{2}\times\dfrac{1}{2}$

$=\dfrac{5}{4}$

$A$ and

$B$ are two independent events such that

$P(A) = \dfrac{1}{2}$ and

$P(B) = \dfrac{2}{3}$ , then

$P((A \cup B) (A\cup \overline{B})(\overline{A} \cup B))$ has the value equal to

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{2}{3}$

In a binomial distribution, the mean is

$\dfrac {2}{3}$ and the variance is

$\dfrac {5}{9}$ . What is the probability that

$X = 2$ ?

0%
$\dfrac {5}{36}$
0%
$\dfrac {25}{36}$
0%
$\dfrac {25}{216}$
0%
$\dfrac {25}{54}$

Explanation

We have

$\mu=np=\dfrac{2}{3}$ and

$\sigma^2=np(1-p)=\dfrac{5}{9}$ .

This give

$p=\dfrac{1}{6}$ and

$n=4$ .

Also,

$P(X=r)=^nC_rp^r(1-p)^{n-r}\implies P(X=2)=^4C_2\left(\dfrac{1}{6}\right)^2\left(\dfrac{5}{6}\right)^2=\dfrac{25}{216}$

If the mean and variance of a binomial variate

$x$ are

$8$ and

$4$ respectively then

$P\left( {X < 3} \right)=$

0%
$\frac{{137}}{{{2^{16}}}}$
0%
$\frac{{697}}{{{2^{16}}}}$
0%
$\frac{{265}}{{{2^{16}}}}$
0%
$\frac{{265}}{{{2^{15}}}}$

Explanation

Solution -

Mean =

$rp=8$

variance =

$rpq= 4$

$q= \dfrac{1}{2}$

$p=\dfrac{1}{2}$

$n=16$ .

$P(X < 3)= P(X=0)+P(X=1)+P(X=2)$

$=\, ^{16}C_{0}\left ( \dfrac{1}{2} \right )\left ( \dfrac{1}{2} \right )^{16}+^{16}C_{p}\left ( \dfrac{1}{2} \right )^{1}\left ( \dfrac{1}{2} \right )^{15}+^{16}C_{2}\left ( \dfrac{1}{2} \right )^{2}\left ( \dfrac{1}{2} \right )^{14}$

$=(^{16}C_{0}+^{16}C_{1}+^{16}C_{2})\left ( \dfrac{1}{2} \right )^{16}= \dfrac{137}{2^{16}}$

A is correct.

1130810_1101820_ans_820ecec605a74fcf9aedf3047e9092bb.jpg

If a random variable X has a poisson distribution such that P(X =1)=P(X=2), its mean and variance are

0%
1,1
0%
2,2
0%
2, $\sqrt{3}$
0%
2,4

An unbiased coin is tossed

$10$ times. By using binomial distribution, find the probability of getting at least

$3$ heads.

0%
$\dfrac{110}{128}$ .
0%
$\dfrac{51}{63}$ .
0%
$\dfrac{121}{128}$ .
0%
$\dfrac{150}{176}$ .

Explanation

= 1 - { Pr(no head) + Pr(1 head ) + Pr(2 heads)

$= 1 - { C(10, 0) (1/2)^{10} + C(10, 1) (1/2)^{10} + C(10, 2)(1/2)^{10} }$

$= 1 - { (1/2)^{10} + 10 (1/2)^{10} + 45 (1/2)^{10} }$

$= 1 - { 56/(2)^{10} } = 1 - (56/1024) = 968/1024$

$= \dfrac{121}{128}$

In a poisson distribution, the probability of

$'0'$ Success is

$10\%$ . The mean of the distribution is equal to?

0%
$\log_{10}e$
0%
$\log_e10$
0%
$0$
0%
$0.1$

Six coins are tossed

$6400$ times. The probability of getting

$6$ heads

$x$ times using poison distribution is

0%
$6400{e^{ - x}}$
0%
$\frac{{6400{e^{ - x}}}}{{x!}}$
0%
$\frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$
0%
${e^{ - 100}}$

Explanation

Therefore, the probability of getting

$6$ heads with

$6$ coins

$= {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{64}} = P\left( { > {a_y}} \right)$
Then,

$\eta P = 6400 \times \frac{1}{{64}} = 100 = m\left( {{a_y}} \right)$
So, by poison's law,

$P\left( {x = n} \right) = \frac{{{e^{ - m}}{m^x}}}{{x!}}$

$= \frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$

A random variable

$X$ is binomial distributed with mean

$12$ and variance is

$8$ . Find the parameter of the distribution are

0%
$18,\dfrac{1}{3}$
0%
$36,\dfrac{1}{3}$
0%
$36,\dfrac{2}{3}$
0%
$18,\dfrac{2}{3}$

Explanation

Solution -

Mean = np = 12

Variance = npq = 8

then q = 1 - p

$\dfrac{npq}{np}=\dfrac{8}{12}$

$q=\dfrac{2}{3}$

$p=1-\dfrac{2}{3}=\dfrac{1}{3}$

$n\times \dfrac{1}{3}=12$

$n=36$

$36,\dfrac{1}{3}$

B is correct

1130802_1101672_ans_532ca4312e114b74ab518ff98e74f8c4.jpg

X and Y are independent binomial variates

$A\left( {5,\dfrac{1}{2}} \right)$ and

$B\left( {7,\dfrac{1}{2}} \right)$ then

$P\left( {X + Y = 3} \right)$

0%
$\dfrac{{45}}{{1024}}$
0%
$\dfrac{{55}}{{1024}}$
0%
$\dfrac{{65}}{{1024}}$
0%
$\dfrac{{60}}{{1024}}$

Explanation

Solution -

$n_{A} = 5 \, P_{A} = q_{A} = \dfrac{1}{2} \, n_{B} = r_{B} = \dfrac{1}{2}$

$P(x+y = 3) = p(x = 0)P(y = 3 )+ p(x=1)P(y=2)$

$+p(x=2)p(y=1)+p(x=3)p(y = 0)$

$^{5}C_{0}\,^{7}C_{3}+^{5}C_{1} \, ^{5}C_{2} ^{7}C_{1}+^{5}C_{3}\,^{7}C_{0}(\dfrac{1}{2})^{7+5}$

$= \dfrac{35+105+70+10}{2^{12}}$

$= \dfrac{220}{2^{12}}$

$= \dfrac{55}{1.24}$

B is correct

1130848_1103795_ans_54915e68ad604782b980b900d1a53b83.jpg

The mean and standard deviation of a binomial variate

$X$ are

$4$ and

$\sqrt 3$ respectively. Then

$P\left( {X \geq 1} \right) =$

0%
$1 - {\left( {\dfrac{1}{4}} \right)^{16}}$
0%
$1 - {\left( {\dfrac{3}{4}} \right)^{16}}$
0%
$1 - {\left( {\dfrac{2}{3}} \right)^{16}}$
0%
` $1 - {\left( {\dfrac{1}{3}} \right)^{16}}$

Explanation

For Binomial distribution, Mean

$(\mu) =np$ and Standard Deviation

$(\sigma)=\sqrt{np(1-p)}$

According to question, Mean = 4 and Variance =

$\sqrt3$

Thus,

$np=4.....(1)\quad \sqrt{np(1-p)}=\sqrt3\Rightarrow np(1-p)=3.......(2)$

Putting the value of np from first equation in the second equation.

$4\times(1-p)=3\Rightarrow1-p=\dfrac{3}{4}\Rightarrow p=\dfrac{1}{4}$

Now, putting the value of p in equation(1),

$n\times\dfrac{1}{4}=4\Rightarrow n=16$

$\because \Sigma P(X_i)=1$

$\therefore P(X\ge1)=1-P(X<1)=1-P(X=0)=1-{^{16}C_0p^0(1-p)^{16-0}}$

$=1-\dfrac{16!}{0!(16-0)!}\times\left(\dfrac{3}{4}\right)^0\times\left(\dfrac{1}{4}\right)^{16}=1-\left(\dfrac{1}{4}\right)^{16}$

On a normal standard die one of the

$21$ dots from any one of the six faces is removed at random with each dot equally likely to be chosen. The die is then rolled. The probability that the top face has an odd number of dots is

0%
$\dfrac {5}{11}$
0%
$\dfrac {5}{12}$
0%
$\dfrac {11}{21}$
0%
$\dfrac {6}{11}$

In a binomial distribution with

$n=4$ , if

$2P(X=3)=3P(X=2)$ , then value of p is?

0%
$\dfrac{9}{13}$
0%
$\dfrac{4}{13}$
0%
$\dfrac{6}{13}$
0%
$\dfrac{7}{13}$

Explanation

$P(X=3) =\space ^4C_3 p^3 (1-p)= 4 p^3 (1-p)$

$P(X=2) = \space ^4C_2 p^2 (1-p)^2 = 6 p^2 (1-p)^2$

$8p^3 (1-p) = 18p^2(1-p)^2$

$4p = 9(1-p)$

$13p = 9$

$p = \cfrac{9}{13}$

In a Binomial distribution, if mean is

$4026$ and variance is

$2013$ , find probability of failure.

0%
$\frac{1}{2}$
0%
$\frac{1}{3}$
0%
$\frac{1}{4}$
0%
$\frac{1}{6}$

Explanation

mean of Binomial distribution

$=4026$

Variance of Binomial distribution

$=2013$

If probability of success if

$P\ 4$ jailme

$\therefore \ np=4026,\ np(1-p)=2013$

$\Rightarrow \ \dfrac {np (1-p)}{np}=\dfrac {2013}{4026}=\dfrac {1}{2}$

$\Rightarrow \ 2(1-b)=1\ \Rightarrow \ 2-2b=1$

$\Rightarrow \ -2b=1-1$

$\Rightarrow \ p=1/2$

Probability of jailme

$=1p=1-1/2=1/2$

Option

$(A)$ is correct

If in

$6$ trials,

$X$ is a binomial variate which follows the relation

$9P(X=4)=P(X=2)$ , then what is the probability of success ?

0%
$3/4$
0%
$1/4$
0%
$3/8$
0%
$1/8$

Explanation

$9\times ^{ 6 }{ C }_{ 4 }{ P }^{ 4 }{ \left( 1-p \right) }^{ 2 }=^{ 6 }{ C }_{ 4 }{ P }^{ 4 }{ \left( 1-p \right) }^{ 2 }$

$\Rightarrow 9{ P }^{ 2 }={ \left( 1-p \right) }^{ 2 }$

$\Rightarrow \cfrac { P }{ 1-p } =3\Rightarrow P=3-3p$

$=4p=3\Rightarrow p=3/4$

There are 4 defective items in a lot consisting of 10 items. From this lot, we select 5 items at random, The probability that there will be 2 defective items among them is -

0%
$\dfrac{1}{2}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{5}{21}$
0%
$\dfrac{10}{21}$

Which one is not a requirement of a binomial distribution?

0%
There are 2 outcomes for each trial
0%
There is a fixed number of trials
0%
The outcomes must be dependent on each other
0%
The probability of success must be the same for all the trials

A examinations consists of

$8$ questions in each of which one of the

$5$ alternatives is the correct one. On the assumption that a candidate who has done no preparatory work, chooses for each questions any one of the five alternatives with equal probability, then the probability that he gets more than one correct answer is equal to:

0%
${\left( {0.8} \right)^8}$
0%
$3{\left( {0.8} \right)^8}$
0%
$1-{\left( {0.8} \right)^8}$
0%
$1-3{\left( {0.8} \right)^8}$

Explanation

Probability of an answers to be correct

$=\dfrac{1}{5}=0.2$

Probability of an answers not to be correct

$=1-0.2$

$=0.8$

Probability

$\left(more\ than\ 1\ correct\right)=1-P\left(0\ correct\right)-P\left(1\ correct\right)$

$=1-^{8}{C}_{0}{\left(0.8\right)}^{8}-^{8}{C}_{1}\left(0.2\right) \left(0.8\right)$

$=1-{\left(0.8\right)}^{8}-1.6{\left(0.8\right)}^{7}$

$=1-{\left(0.8\right)}^{8}-2{\left(0.8\right)}^{8}$

$=1-3{\left(0.8\right)}^{8}$

$D$ is coorect.

Which one is not a requirement of a Binomial distribution?

0%
There are $2$ outcomes for each trial
0%
There is a fixed number of trials
0%
The outcomes must be dependent on each other
0%
The probability of success must be the same for all the trials

A contest consists of prediciting the results (win,draw or defeat) of 10 footballs matches. The probability that one entry contains at least 5 correct answers is

0%
$\dfrac{12585}{3^{10}}$
0%
$\dfrac{12385}{3^{10}}$
0%
$\dfrac{9335}{3^{10}}$
0%
$\dfrac{12096}{3^{10}}$

Explanation

P(correct)=

$\dfrac{1}{3}$

P(wrong)=

$\dfrac{2}{3}$

here

$n=5$

$=^{10}C_5 \left ( \dfrac{1}{3} \right )^5\left ( \dfrac{2}{3} \right )^5$

$=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 2^5}{(5\times 4\times 3\times 2)(5\times 4\times 3\times 2)\times 3^5\times 3^5}$

$=\dfrac {252\times 2^5}{3^{10}}$

$=\dfrac{12096}{3^{10}}$

Consider 5 independent Bernoulli's trials each with probability of success p. If the probability of at
least one failure is greater than or equal to

$\frac { 31 }{ 32 } ,$ then p lies in the interval :-

0%
$\left[ 0,\frac { 1 }{ 2 } \right]$
0%
$\left[ \frac { 11 }{ 12 } ,1 \right]$
0%
$\left[ \frac { 1 }{ 2 } ,\frac { 3 }{ 4 } \right]$
0%
$\left[ \frac { 3 }{ 4 } ,\frac { 11 }{ 12 } \right]$

The least number of times a fair coin must be tossed so that the probability of getting atleast one head is

$0.8$ , is

0%
$7$
0%
$6$
0%
$5$
0%
$3$

Explanation

In a single toss, or either get a head or a tail

Probability of getting a head in a single toss =

$\frac{1}{2}$

Probability of getting no head in a single toss =1/2

Probability of getting no head in n toss

$= (\frac{1}{2})^{n}$

Probability of getting atleast one head in n tosses

= 1- Probability of getting no heads in a tosses.

$= 1-(\frac{1}{2})^{n}$

and as per question ,

$1-(\frac{1}{2})^{n}=0.8$

$1-0.8=(\frac{1}{2})^{n}$

$\Rightarrow 0.2=(\frac{1}{2})^{n}$

$\Rightarrow 0.2 = 2^{\frac{1}{2}}$

$\Rightarrow 2^{n}=\frac{1}{0.2}$

$\Rightarrow 2^{n}=5$

$2^{1}=2 < 5$

$2^{2}= 4 < 5$

$2^{3}= 8 > 5$

$\therefore$ The least number of limits, n=3.

Which of the following is not true regarding the normal distribution?

0%
the point of inflecting are at $X = \mu \pm \sigma$
0%
skewness is zero
0%
maximum heigth of the curve is $\dfrac{1}{\sqrt{2\pi}}$
0%
mean $=$ media $=$ mode

Explanation

$\left(i\right)$ Since

$f\left(x\right)$ is a nonzero function we may divide both sides of the equation by this function. From this

it is easy to see that the inflection points occur where

$X =\mu\pm\sigma$ . In other words the inflection points are

located one standard deviation above the mean and one standard deviation below the mean

$\left(ii\right)$ The skewness for perfect normal distribution is

$0.0$ . But if sample is greater than

$100$ and less

than

$200$ , the acceptable absolute skewness value is

$1.0$ . However for large sample size

$n$ greater than

$200$ ,

the absolute value for acceptable skewness is

$1.5$ .

$\left(iii\right)$ The area under the normal curve is equal to

$1.0$ . Normal distributions are denser in the center and

less dense in the tails. Normal distributions are defined by two parameters, the mean

$\left(\mu\right)$ and the standard

deviation

$\left(\sigma\right)$ .

$68\%$ of the area of a normal distribution is within one standard deviation of the mean.

$\left(iv\right)$ The mean, median, and mode of a normal distribution are equal. The area under the normal curve is equal to 1.0.

Normal distributions are denser in the center and less dense in the tails. Normal distributions are defined by two

parameters, the mean

$\left(\mu\right)$ and the standard deviation

$\left(\sigma\right)$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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