Get Answers to all your Questions

Filter By

Clear Filter

All Questions

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

View Full Answer(11)

Posted by

kpvalli2725@gmail.com

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 $\cap$ B)= $\frac{1}{4}$

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

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

P (A) + P (B) -2 P (A $\cap$ B) - P(B $\cap$ C)-P(C $\cap$ A) = $\frac{3}{8}$

P(A $\cap$ B $\cap$ C)= $\frac{1}{16}$

P(A $\cap$ B $\cap$ C)= $\frac{3}{8}+\frac{1}{16}=\frac{4}{16}$

View Full Answer(1)

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

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)

View Full Answer(1)

Posted by

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

The sum of the deviation of an observation from their mean is equal to zero. i.e. $\sum^{n}_{i=1}\left ( x_i-\bar x \right )=0$ .
The sum of the square of the deviation from the mean is minimum, i.e. $\sum^{n}_{i=1}\left ( x_i-\bar x \right )^2\;\text{is least.}$
The mean is affected accordingly if the observations are given a mathematical treatment i.e. addition, subtraction multiplication by a constant term.
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 .
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, $\\d_1=\bar x_2-\bar x\;\;\;\text{and}\;\;\;d_2=\bar x_ 1-\bar x$

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

$\bar x_1,\;\;\bar x_2$ are 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)

View Full Answer(1)

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

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 ,..., x_n 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

$n+m = 18$

View Full Answer(1)

Posted by

Kuldeep Maurya

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

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

View Full Answer(1)

Posted by

Ritika Jonwal

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

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 ,..., x_n 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)

View Full Answer(1)

Posted by

vishal kumar

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)

View Full Answer(1)

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

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, ……, x_n, 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 ,..., x_n is given by

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

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

View Full Answer(1)

Posted by

Ritika Jonwal

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)

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Top Schools

Products & Resources

NCERT Study Material

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

A random variable X has the following probability distribution: Then is equal to: Option: 1 Option: 2 Option: 3 Option: 4

Posted by

kpvalli2725@gmail.com

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) and P(All the three events occur simultaneously) Then the probability that at least one of the events occurs, is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

Let A and B be two independent events such that and . Then, which of the following is TRUE? Option: 1 Option: 2 Option: 3 Option: 4

Posted by

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 and If the new mean and new s.d. become half of their original values, then q is equal to :Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

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

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

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 $4.02$

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 $0.02$

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