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A random variable X has the following probability distribution:          X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5 P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2} Then P(X>2) is equal to: 
Option: 1 \frac{7}{12}
Option: 2 \frac{23}{36}
Option: 3 \frac{1}{36}
Option: 4 \frac{1}{6}
 

Option 2

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For three events A, B and C, P(Exactly one of A or B occurs) =P(Exactly one of B or C occurs)
=P(Exactly one of C or A occurs)  =\frac{1}{4}   and P(All the three events occur
simultaneously)  =\frac{1}{16} Then the probability that at least one of the events occurs, is :  
Option: 1 \frac{7}{16}

Option: 2 \frac{7}{64}

Option: 3 \frac{3}{16}

Option: 4 \frac{7}{32}
 

P (exactly one of A or B) =P(exactly B or C)

=P (exactly one of A or C) = \frac{1}{4}

P(A) + P (B) - 2P (A\capB)=\frac{1}{4}

P(B) +P (C) - 2P (B\capC)=\frac{1}{4}

P(C)+P(A)-2P (C\capA)=\frac{1}{4}

P (A) + P (B) -2 P (A\capB) - P(B\capC)-P(C\capA) =\frac{3}{8}

P(A\capB\capC)=\frac{1}{16}

P(A\capB\capC)=\frac{3}{8}+\frac{1}{16}=\frac{4}{16}

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Let A and B be two independent events such that P(A)=\frac{1}{3} and P(B)=\frac{1}{6}. Then, which of the following is TRUE?
Option: 1 P(B/A) = \frac{1}{2}
Option: 2 P\left ( A/B' \right )=\frac{1}{3}
Option: 3 P\left ( A/B \right )=\frac{2}{3}
Option: 4 P\left ( A'/B' \right )=\frac{1}{3}
 

A & B are independent events

So, A & B' are also independent events, and hence

P\left(\frac{A}{B^{\prime}}\right)=P(A)=\frac{1}{3}

Correct Option (2)

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Kuldeep Maurya

The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where p\neq 0 and q\neq 0. If the new mean and new s.d. become half of their original values, then q is equal to :
Option: 1 -20
Option: 2 -5
Option: 3 10  
Option: 4 -10

 

 

Some Important Point Regarding Statistics -

Some Important Point Regarding Statistics

  1. The sum of the deviation of an observation from their mean is equal to zero. i.e. \sum^{n}_{i=1}\left ( x_i-\bar x \right )=0.
  2. The sum of the square of the deviation from the mean is minimum, i.e. \sum^{n}_{i=1}\left ( x_i-\bar x \right )^2\;\text{is least.}
  3. The mean is affected accordingly if the observations are given a mathematical treatment i.e. addition, subtraction multiplication by a constant term.
  4. If set of n1 observations has mean \bar x_1 and set of n2 observations has mean \bar x_2, then their combined mean is \frac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2} and the .
  5. If set of  n1 observations has mean \bar x_1 and set of n2 observations has mean \bar x_2  then their combined variance is given by

            \sigma ^{2}= \frac{n_{1}\left ( \sigma _{1}^{2}+d_{1}^{2} \right )+n_{2}\left ( \sigma _{2}^{2}+d_{2}^{2} \right )}{n_{1}+n_{2}}

          where, \\d_1=\bar x_2-\bar x\;\;\;\text{and}\;\;\;d_2=\bar x_ 1-\bar x 

          \\ {\sigma_{1}^{2}=\frac{1}{n_{1}} \sum_{i=1}^{n_{1}}\left(x_{1 i}-\overline{x_{1}}\right)^{2}} \\\\ {\sigma_{2}^{2}=\frac{1}{n_{2}} \sum_{j=1}^{n_{2}}\left(x_{2 j}-\overline{x_{2}}\right)^{2}}

          \bar x_1,\;\;\bar x_2are the means and \sigma_1,\;\;\sigma_2  are the standard deviations of two series.

-

\text{old series} \Rightarrow x_{old}=x_1,x_2,x_3\ldots\ldots x_n

Old Mean

\overline x_{old}=\frac{\left (x_1+x_2+x_3\ldots\ldots x_n \right ) }{n}=20

Old Standard Deviation

\sigma^{2}_{old}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

If each observation is multiplied with p & then q is subtracted

\text{new series} \Rightarrow x_{new}=px_1-q,\;px_2-q,\;px_3-q\ldots\ldots \;px_n-q

New mean is the half of old mean

\\ \overline x_{new}=\frac{p\left (x_1+x_2+x_3\ldots\ldots x_n \right )-nq}{n} =10\\\\ \overline x_{new}=p\;\overline x_{old}-q=10\\\\ \overline x_{new}=20p-q=10\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i) 

and new standard deviations is the half of old mean

\sigma^{2}_{new}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{new}-\bar{x}_{new}\right)^{2}

\sigma^{2}_{new}=\frac{1}{n} \sum_{i=1}^{n}\left[(px_{i}-q)-(p\bar{x}_{old}-q)\right]^{2}

\sigma^{2}_{new}=\frac{1}{n} \sum_{i=1}^{n}\left(px_{i}-p\bar{x}_{old}\right)^{2}

\sigma^{2}_{new}=\frac{1}{n}p^2 \sum_{i=1}^{n}\left(x_{i}-\bar{x}_{old}\right)^{2}

\sigma_{new}=|p|\sigma_{old}

1=2|p|

p=\pm0.5

Case 1 p=0.5

Using equation 1

we get

\\20p-q=10\\10-q=10\\q=0

Case 1 p=-0.5

Using equation 1

we get

\\20p-q=10\\-10-q=10\\q=-20

Correct Option (1)

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If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m+n is equal to _____.
Option: 1 12
Option: 2 18
Option: 3 16
Option: 4 20

 

 

Dispersion (Variance and Standard Deviation) -

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity which leads to a proper measure of dispersion. 

The variance of n observations x1 , x2 ,..., xn is given by

\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

Standard Deviation

The standard deviation is a number that measures how far data values are from their mean.

The positive square-root of the variance is called standard deviation. The standard deviation, usually denoted by σ  and it is given by

\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}

-

\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}

variance of the first n natural numbers is 10

n = 1,2,3,4........n

\sum x_i={1+2+3.....n}=\frac{n(n+1)}{2}

\sum x_i^2={1^2+2^2+3^2.....n^2}=\frac{n(n+1)(2n+1)}{6}

Using variance formula

\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}

\sigma ^{2}= \left ( \frac{n(n+1)(2n+1)}{6n} \right )-\left ( \frac{n(n+1)}{2n} \right )^{2}

\sigma ^{2}= \left ( \frac{(n+1)(2n+1)}{6} \right )-\left ( \frac{(n+1)}{2} \right )^{2}

\sigma ^{2}=\frac{n+1}{2}\left [ \left ( \frac{2n+1}{3} \right )-\left ( \frac{n+1}{2} \right ) \right ]

\sigma ^{2}=\frac{n+1}{2} \left ( \frac{4n+2-3n-3}{6} \right )

\\\sigma _n=\frac{n^2-1}{12}=10 \\

n^2=121\;\;\;\;\Rightarrow\;\;\;\;n=11

 

variance of the first m even natural numbers is 16

N = 2,4,6,8........2m

N = 2(1,2,3,4........m)

\sum x_j=2{(1+2+3.....m)}=2\times\frac{m(m+1)}{2}=m(m+1)

\sum x_j^2=2^2{(1^2+2^2+3^2.....m^2)}=4\times\frac{m(m+1)(2m+1)}{6}

Using variance formula

\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}

\sigma ^{2}= \left ( \frac{2m(m+1)(2m+1)}{3m} \right )-\left ( \frac{m(m+1)}{m} \right )^{2}

\sigma ^{2}= \left ( \frac{2(m+1)(2m+1)}{3} \right )-\left (m+1 \right )^{2}

\sigma ^{2}=\frac{m+1}{3} \left ( 4m+2-3m-3 \right )

\\ \sigma _{m(even)} = \frac{m^2-1}{3}=16

m^2=49\;\;\;\Rightarrow\;\;\;m=7

Hence

                n+m = 18

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An unbiased coin is tossed 5 times.Suppose that a variable X is assigned the value k when consecutive heads are obtained for k = 3,4,5, otherwise X takes the value -1. Then the expected value of X, is :
Option: 1 \frac{1}{8}


Option: 2 \frac{3}{16}


Option: 3 -\frac{1}{8}


Option: 4 -\frac{3}{16}

An unbiased coin is tossed 5 times, so total number of outcome is 25 = 32.

Now, the probability for getting number of heads occurring simultaneously

\\\mathrm{for\;k=0}\\\mathrm{E(k)=\{T T T T T\}=1}\\\mathrm{P(k)=\frac{1}{32}}

\\\mathrm{for\;k=1}\\\mathrm{E(k)=\{ HTTTT,THTTT,TTHTT,TTTHT,TTTTH,HTHTT,HTTHT,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;HTTTH,THTHT,THTTH,TTHTH,HTHTH\} =12}\\\mathrm{P(k)=\frac{12}{32}}

\\\mathrm{for\;k=2}\\\mathrm{E(k)=\{ { HHTTT,THHTT,TTHHT,TTTHH,HHTHT,HHTTH },}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; { THHTH,HTHHT,HTTHH,THTHH,HHTHH }\} =11}\\\mathrm{P(k)=\frac{11}{32}}

\\\mathrm{for\;k=3}\\\mathrm{E(k)=\{ HHHTT,THHHT,TTHHH,HHHTH,HTHHH\} =5}\\\mathrm{P(k)=\frac{5}{32}}

\\\mathrm{for\;k=4}\\\mathrm{E(k)=\{ \mathrm{HHHHT,THHHH}\} =2}\\\mathrm{P(k)=\frac{2}{32}}

\\\mathrm{for\;k=5}\\\mathrm{E(k)=\{ \mathrm{HHHHH}\} =1}\\\mathrm{P(k)=\frac{1}{32}}

\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{K} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{k}) & \frac{1}{32} & \frac{12}{32} & \frac{11}{32} & \frac{5}{32} & \frac{2}{32} & \frac{1}{32} \\ \hline \end{array}

\text { For } \mathrm{k}=0,1,2 ;\mathrm{X}=-1 \text { and for } \mathrm{k}=3,4,5; \mathrm{X}=\mathrm{k} \text { . }

Now expected value is

\\\sum X \times P(k)=(-1) \times \frac{1}{32}+(-1) \times \frac{12}{32}+(-1) \times \frac{11}{32} +3 \times \frac{5}{32}+4 \times \frac{2}{32}+5 \times \frac{1}{32}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{8}

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The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is :
Option: 1 4.01
Option: 2 3.99
Option: 3 3.98
Option: 4 4.02

 

 

Dispersion (Variance and Standard Deviation) -

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity which leads to a proper measure of dispersion. 

The variance of n observations x1 , x2 ,..., xn is given by

\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

Standard Deviation

The standard deviation is a number that measures how far data values are from their mean.

The positive square-root of the variance is called standard deviation. The standard deviation, usually denoted by σ  and it is given by

\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}


Variance and Standard Deviation of a Discrete Frequency Distribution

The given discrete frequency distribution be

         \begin{matrix} x &: &x_1, & x_2, & x_3, &\ldots &x _n & \\ & & & & & & & \\ f& : & f_1, &f_2, &f_3, &\ldots & f_n & \end{matrix}

\\\text{In this case, Variance}\;\left ( \sigma^2 \right ) \;\;\;=\frac{1}{N} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}\\\\\text{and, Standard Deviation}(\sigma)\;=\sqrt{\frac{1}{N} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}}\\\text{where, N}=\sum_{i=1}^{n} f_{i}

 

Variance and Standard deviation of a continuous frequency distribution 

The formula for variance and standard deviation are the same as in the case of discrete frequency distribution. Here, x_i is the mid point of each class.

 

Another formula for Standard Deviation

\\\text{Variance }\left ( \sigma^2 \right )=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(x_{i}^{2}+\bar{x}^{2}-2 \bar{x} x_{i}\right)\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\sum_{i=1}^{n} \bar{x}^{2} f_{i}-\sum_{i=1}^{n} 2 \bar{x} f_{i} x_{i}\right]\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2} \sum_{i=1}^{n} f_{i}-2 \bar{x} \sum_{i=1}^{n} x_{i} f_{i}\right]\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2} N-2 \bar{x} \cdot N\bar x\right]

\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2} N-2 \bar{x} \cdot N\bar x\right]\\\\\left [ \because \frac{1}{N} \sum_{i=1}^{n} x_{i} f_{i}=\bar{x} \text { or } \sum_{i=1}^{n} x_{i} f_{i}=\mathrm{N} \bar{x} \right ]\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2}-2 \bar{x}^{2}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i} x_{i}^{2}-\bar{x}^{2}\\\\\text{or}\;\;\sigma^{2}=\frac{1}{N} \sum_{i=1}^{n} f_{i} x_{i}^{2}-\left(\frac{\sum_{i=1}^{n} f_{i} x_{i}}{N}\right)^{2}=\frac{1}{N^{2}}\left[N \sum_{i=1}^{n} f_{i} x_{i}^{2}-\left(\sum_{i=1}^{n} f_{i} x_{i}\right)^{2}\right]

\text{Thus, standard deviation }(\sigma)=\frac{1}{N} \sqrt{N \sum_{i=1}^{n} f_{i} x_{i}^{2}-\left(\sum_{i=1}^{n} f_{i} x_{i}\right)^{2}}



Shortcut method to find variance and standard deviation

The values of x_i  in a discrete distribution or the mid points x_i of different classes in a continuous distribution are large and so the calculation of mean and variance becomes tedious and time consuming.

Here is the shortcut method to find variance and standard deviatio

Let the assumed mean be ‘A’ and the scale be reduced to 1/h times (h being the width of class-intervals). 

Let the step-deviations or the new values be y_i .

\\\text { i.e. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad y_{i}=\frac{x_{i}-\mathrm{A}}{h} \text { or } x_{i}=\mathrm{A}+h y_{i}|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)\\\\\text { We know that } \quad\;\;\;\;\;\bar x= \frac{\sum_{i=1}^{n} f_{i} x_{i}}{N}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(2)

Replacing x_i from (1) in (2), 

 

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\bar{x}=\frac{\sum_{i=1}^{n} f_{i}\left(\mathrm{A}+h y_{i}\right)}{\mathrm{N}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{\mathrm{N}}\left(\sum_{i=1}^{n} f_{i} \mathrm{A}+\sum_{i=1}^{n} h f_{i} y_{i}\right)=\frac{1}{\mathrm{N}}\left(\mathrm{A} \sum_{i=1}^{n} f_{i}+h \sum_{i=1}^{n} f_{i} y_{i}\right)\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\mathrm{A} \cdot \frac{\mathrm{N}}{\mathrm{N}}+h \frac{\sum_{i=1}^{n} f_{i} y_{i}}{\mathrm{N}} \quad\left(\text { because } \sum_{i=1}^{n} f_{i}=\mathrm{N}\right)\\\text { Thus } \quad \bar{x}=\mathrm{A}+h \bar{y}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(3)
Now Variance of the variable x

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\sigma_{x}^{2}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} =\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(\mathrm{A}+h y_{i}-\mathrm{A}-h \bar{y}\right)^{2}\;\;\;\;\;\;\;\;\;\text{(Using (1) and (3))}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} =\frac{1}{N} \sum_{i=1}^{n} f_{i} h^{2}\left(y_{i}-\bar{y}\right)^{2}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} =\frac{h^{2}}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(y_{i}-\bar{y}\right)^{2}=h^{2} \times \text { variance of the variable } y_{i}\\\text { i.e. }\;\;\;\; \quad \sigma_{x}^{2}=h^{2} \sigma_{y}^{2}\;\;\\\text { or }\;\;\;\;\;\; \quad \sigma_{x}=h \sigma_{y}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(4)

From (3) and (4),

\sigma_{x}=\frac{h}{\mathrm{N}} \sqrt{\mathrm{N} \sum_{i=1}^{n} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{n} f_{i} y_{i}\right)^{2}}

-

 

\\\\\sigma_{new}^2=4-\left [\frac{(11-9)}{20} \right ]^2=4-\frac{4}{200}=3.99

Correct Option (2)

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Let A and B be two events such that the probability that exactly one of them occurs is \frac{2}{5} and the probability that A or B occurs is \frac{1}{2}, then the probability pf both of them occur together is :  
Option: 1 0.10
Option: 2 0.20
Option: 3 0.01
Option: 4 0.02

 

\\P(\text{exactly one})=\frac{2}{5}=P(A)+P(B)-2P(A \cap B)\\ \ P(A \cup B)=\frac{1}{2}=P(A)+P(B)-P(A\cap B)\\ \Rightarrow P(A \cap B)=\frac{1}{2}-\frac{2}{5}=\frac{1}{10}

Correct Option (1)

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If the mean and variance of eight numbers 3,7,9,12,13,20,x and y be 10 and 25 respectively, then x\cdot y is equal to _____.
Option: 1 54
Option: 2 60
Option: 3 24
Option: 4 None

 

 

Mean -

Mean of the Ungrouped Data 

If n observations in data are x1, x2, x3, ……, xn, then arithmetic mean \mathit{\bar x} is given by    

\mathit{\bar x}=\frac{x_1+x_2+x_3+\ldots\dots +x_n }{n}=\frac{1}{n}\sum^{n}_{i=1}x_i

-

 

 

Dispersion (Variance and Standard Deviation) -

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity which leads to a proper measure of dispersion. 

The variance of n observations x1 , x2 ,..., xn is given by

\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

-

\\\text{Mean}=\bar x=\frac{3+ 7+ 9+ 12+ 13+ 20 +x+ y}{8}=10\\\Rightarrow x+y=16\\\text{variance}\;\;\sigma^2=\frac{\sum (x_i)^2}{8}-(\bar x)^2\\\Rightarrow \frac{9 +49 +81+144+ 169 +400 +x^2 +y^2}{8}-100=25\\\Rightarrow x^2+y^2=148\\(x+y)^2=x^2+y^2+2xy\\xy=54

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In a workshop, there are five machines and the probability of any one of them to be out of service on a day is \frac{1}{4}. If the probability that at most two machines will be out of service on the same day is \left ( \frac{3}{4} \right )^{3}k, then k is equal to :
Option: 1 \frac{17}{2}
Option: 2 4
Option: 3 \frac{17}{4}
Option: 4 \frac{17}{8}

 

Required probability = when no. the machine has fault + when only one machine has fault + when only two machines have the fault.

\\\text {P(R)}=\;^{5} \mathrm{C}_{0}\left(\frac{3}{4}\right)^{5}+^{5} \mathrm{C}_{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{4}+^{5} \mathrm{C}_{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3} \\\text {P(R)}=\frac{243}{1024}+\frac{405}{1024}+\frac{270}{1024}=\frac{918}{1024}\\\text {P(R)}=\frac{459}{512}\\\text {P(R)}=\frac{27 \times 17}{64 \times 8} =\left(\frac{3}{4}\right)^{3} \times \mathrm{k}=\left(\frac{3}{4}\right)^{3} \times \frac{17}{8} \\\therefore \mathrm{k}=\frac{17}{8}Correct Option (4)

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