Statistics
Let the mean and the variance of 6 observations $a, b, 68,44,48,60$ be $55$ and $194$, respectively. If $a>b$, then $a+3 b$ is
Let M denote the median of the following frequency distribution
| Class | 0 - 4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 |
|---|---|---|---|---|---|
| Frequency | 3 | 9 | 10 | 8 | 6 |
Then 20M is equal to :
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$, then the variance of the first four observations in equal to
Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{N}$ and $\mathrm{a}< \mathrm{b}< \mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $9,25, a, b, c$ be 18, 4 and $\frac{136}{5}$, respectively. Then $2 a+b-c$ is equal to ________
Explanation:
$\begin{aligned} & a, b, c \in N \\ & a< b < c \\ & \text { Mean }=18 \\ & \frac{9+25+a+b+c}{5}=18 \\ & 34+a+b+c=90 \\ & a+b+c=56 \end{aligned}$
$\begin{aligned} & \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\ & 9+7+|a-18|+|b-18|+|c-18|=20 \\ & |a-18|+|b-18|+|c-18|=4 \\ & \frac{136}{5}=\frac{706+a^2+b^2+c^2}{5}-(18)^2 \\ & \Rightarrow 136=706+a^2+b^2+c^2-1620 \\ & \Rightarrow a^2+b^2+c^2=1050 \\ & \text { Consider } a<19 < b< c \\ & \text { Solving } a=17, b=19, c=20 \\ & 2 a+b-c \\ & 34+19-20 \\ & =33 \end{aligned}$
Let the mean and the standard deviation of the probability distribution
| $\mathrm{X}$ | $\alpha$ | 1 | 0 | $-$3 |
|---|---|---|---|---|
| $\mathrm{P(X)}$ | $\frac{1}{3}$ | $\mathrm{K}$ | $\frac{1}{6}$ | $\frac{1}{4}$ |
be $\mu$ and $\sigma$, respectively. If $\sigma-\mu=2$, then $\sigma+\mu$ is equal to ________.
Explanation:
Mean $(\mu)=\Sigma x_i P\left(x_i\right)$
Standard deviation $(\sigma)=\sqrt{\left(\Sigma x_i^2 P\left(x_i\right)\right)-\mu^2}$
$\begin{aligned} & \Rightarrow \quad \mu=\frac{1}{3} \alpha+K-\frac{3}{4} \\ & \sigma=\sqrt{\left(\frac{1}{3} \alpha^2+K+0+\frac{9}{4}\right)-\left(\frac{1}{3} \alpha+K-\frac{3}{4}\right)^2} \\ & \because \Sigma P_i=1 \Rightarrow \frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1 \\ & \Rightarrow \quad K=\frac{1}{4} \Rightarrow \mu=\frac{1}{3} \alpha-\frac{1}{2} \\ & \because \sigma-\mu=2 \\ & \sigma^2=(\mu+2)^2 \end{aligned}$
$\begin{aligned} & \frac{1}{3} \alpha^2+\frac{5}{2}-\mu^2=(\mu+2)^2 \\ & \frac{1}{3} \alpha^2+\frac{5}{2}=\left(\frac{1}{3} \alpha-\frac{1}{2}\right)^2+\left(\frac{1}{3} \alpha+\frac{3}{2}\right)^2 \\ & \Rightarrow \alpha=0,6 \end{aligned}$
$\begin{array}{ll} \text { If } \alpha=0, K=\frac{1}{4} & \text { If } \alpha=6, K=\frac{1}{4} \\ \mu=-\frac{1}{2}, \sigma=\frac{3}{2} & \mu=\frac{3}{2}, \sigma=\frac{7}{2} \\ \sigma+\mu=1 & \sigma+\mu=5 \end{array}$
Both (1) and (5) are correct but according to NTA (5) is correct
The variance $\sigma^2$ of the data
| $x_i$ | 0 | 1 | 5 | 6 | 10 | 12 | 17 |
|---|---|---|---|---|---|---|---|
| $f_i$ | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
is _________.
Explanation:
| $\mathrm{x_i}$ | $\mathrm{f_i}$ | $\mathrm{f_i x_i}$ | $\mathrm{f_i x^2_i}$ |
|---|---|---|---|
| 0 | 3 | 0 | 0 |
| 1 | 2 | 2 | 2 |
| 5 | 3 | 15 | 75 |
| 6 | 2 | 12 | 72 |
| 10 | 6 | 60 | 600 |
| 12 | 3 | 36 | 432 |
| 17 | 3 | 51 | 867 |
| $\Sigma \mathrm{f}_{\mathrm{i}}=22$ | $\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=2048$ |
$\begin{aligned} & \therefore \quad \Sigma \mathrm{f}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=176 \\ & \text { So } \overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{176}{22}=8 \\ & \text { for } \sigma^2=\frac{1}{\mathrm{~N}} \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^2-(\overline{\mathrm{x}})^2 \\ & =\frac{1}{22} \times 2048-(8)^2 \\ & =93.090964 \\ & =29.0909 \\ & \end{aligned}$
If the mean and variance of the data $65,68,58,44,48,45,60, \alpha, \beta, 60$ where $\alpha> \beta$, are 56 and 66.2 respectively, then $\alpha^2+\beta^2$ is equal to _________.
Explanation:
$\begin{aligned} & \overline{\mathrm{x}}=56 \\ & \sigma^2=66.2 \\ & \Rightarrow \frac{\alpha^2+\beta^2+25678}{10}-(56)^2=66.2 \\ & \therefore \alpha^2+\beta^2=6344 \end{aligned}$
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to __________.
Explanation:
Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$
We have
$\mu^{\prime}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}=180$
As per given information correct $\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$
$\Rightarrow \mu(\text { correct mean})=\frac{182}{15}$
Also
$\sigma^{\prime}=\sqrt{\frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295$
Correct $\Sigma \mathrm{x}_{\mathrm{i}}{ }^2=2295-100+144=2339$
$\sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15} $
Required value
$\begin{aligned} & =15\left(\mu+\mu^2+\sigma^2\right) \\ & =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\ & =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\ & =2521 \end{aligned}$
The variance of the following continuous frequency distribution is
| Classinterval | 0-4 | 4-8 | 8-12 | 12-16 |
| Frequency | 2 | 3 | 2 | 1 |
Based on the following statements, choose the correct option.
Statement I The variance of the first $n$ even natural numbers is $\frac{n^2-1}{4}$.
Statement II The difference between the variance of the first 20 even natural numbers and their arithmetic mean is 112 .
The coefficient of variation for the frequency distribution is
$ \begin{array}{|c|l|l|l|} \hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 3 & 1 \\ \hline \boldsymbol{f}_{\boldsymbol{i}} & 1 & 3 & 5 \\ \hline \end{array} $
Mean deviation about the mean for the following data is
$ \begin{array}{llllll} \hline \text { Class Interval } & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \hline \text { Frequency } & 1 & \,2 & \,3 & \,2 & \,1 \\ \hline \end{array} $
If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data, then $m+\sigma^2=$
| $\mathbf{x}$ | 1 | 3 | 5 | 7 | 9 |
| $\mathbf{f}$ | 4 | 24 | 28 | 16 | 8 |
Let the mean of 6 observations $1,2,4,5, \mathrm{x}$ and $\mathrm{y}$ be 5 and their variance be 10 . Then their mean deviation about the mean is equal to :
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of $\mathrm{A}$ and adding 2 to each element of $\mathrm{B}$. Then the sum of the mean and variance of the elements of $\mathrm{C}$ is ___________.
Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution
| ${x_i}$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| ${f_i}$ | $k + 2$ | $2k$ | ${k^2} - 1$ | ${k^2} - 1$ | ${k^2} + 1$ | $k - 3$ |
where $\sum f_{i}=62$. If $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^{2}+\sigma^{2}\right]$ is equal to :
Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to :
The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $\sigma^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13 , then $\sigma^{2}$ is equal to :
Let $9=x_{1} < x_{2} < \ldots < x_{7}$ be in an A.P. with common difference d. If the standard deviation of $x_{1}, x_{2}..., x_{7}$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_{6}$ is equal to :
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :
Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If $\mu$ and $\sigma^2$ represent mean and variance of X, respectively, then $10(\mu^2+\sigma^2)$ is equal to :
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :
Let the six numbers $\mathrm{a_1,a_2,a_3,a_4,a_5,a_6}$, be in A.P. and $\mathrm{a_1+a_3=10}$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then 8$\sigma^2$ is equal to :
The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________
Explanation:
- The initial mean is given by:
$\bar{x}=50$
So, the total sum of the marks initially was:
$\sum x_i = \bar{x} \times n = 50 \times 10 = 500$
- We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:
$\sum x_{i{\text{correct}}} = \sum x_i - 45 - 50 + 20 + 25 = 500 - 45 - 50 + 20 + 25 = 450$
- The initial variance is given as:
$\sigma^2 = 144$
We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:
$\frac{\sum x_i^2}{n} = \sigma^2 + \bar{x}^2 = 144 + 50^2 = 2644$
Then, the sum of the squares of the initial marks is:
$\sum x_i^2 = n \times \frac{\sum x_i^2}{n} = 10 \times 2594 = 26440$
- The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:
$\sum x_{i{\text{correct}}}^2 = \sum x_i^2 - 45^2 - 50^2 + 20^2 + 25^2 = 26400 - 45^2 - 50^2 + 20^2 + 25^2 = 22940$
- Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:
$\sigma_{\text{correct}}^2 = \frac{\sum x_{i{\text{correct}}}^2}{n} - \left(\frac{\sum x{i_{\text{correct}}}}{n}\right)^2 = \frac{22940}{10} - \left(\frac{450}{10}\right)^2 = 2294 - 45^2 = 2294 - 2025 = 269$
Therefore, the correct variance is 269.
Let the mean of the data
| $x$ | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|
| Frequency ($f$) | 4 | 24 | 28 | $\alpha$ | 8 |
be 5. If $m$ and $\sigma^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^{2}}$ is equal to __________
Explanation:
$ \begin{aligned} \sum f_i & =80 \\\\ \text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\ & =\frac{8}{5} \end{aligned} $
$ \begin{aligned} \sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5} \end{aligned} $
So, $ \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8 $
Let the positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_{3}+a_{4}+a_{5}=14$, then $m+n$ is equal to ___________.
Explanation:
$a_2 = r a$,
$a_3 = r^2 a$,
$a_4 = r^3 a$,
$a_5 = r^4 a$.
where $r$ is the common ratio and $a_1$ = $a$ is the first term.
Given that the mean of the series is $\frac{31}{10}$, we get
$\frac{1}{5}(a_1 + a_2 + a_3 + a_4 + a_5) = \frac{31}{10}$
Substituting the terms with the values of $a_1$ and $r$ gives
$\frac{1}{5}(a + r a + r^2 a + r^3 a + r^4 a) = \frac{31}{10}$
Simplifying this gives
$a (1 + r + r^2 + r^3 + r^4) = \frac{31}{2}$
$ \begin{aligned} & \frac{a\left(r^5-1\right)}{r-1}=\frac{31}{2} ......(1) \\\\ & \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}\right)=\frac{31}{40} \cdot 5=\frac{31}{8} \\\\ & \frac{1}{a}\left(\frac{1-\left(\frac{1}{r}\right)^5}{1-\frac{1}{r}}\right)=\frac{31}{8} \\\\ & \text { or } \frac{1}{a}\left(\frac{r^5-1}{r-1}\right) \frac{1}{r^4}=\frac{31}{8} .........(2) \end{aligned} $
From (1) and (2)
$ \frac{1}{a} \cdot \frac{31}{2 a} \cdot \frac{1}{r^4}=\frac{31}{8} $
$ a r^2=2 $
From (1)
$ \begin{aligned} & \frac{2}{r^2}\left(\frac{r^5-1}{r-1}\right)=\frac{31}{2} \\\\ & \frac{1+r+r^2+r^3+r^4}{r^2}=\frac{31}{4} \\\\ & \left(r^2+\frac{1}{r^2}\right)+\left(r+\frac{1}{r}\right)=\frac{27}{4} \\\\ & t^2-2+t=\frac{27}{4} \end{aligned} $
$ \begin{aligned} &4 t^2+4 t-35=0\\\\ &4 t^2+14 t-10 t-35=0\\\\ &(2 t-5)(2 t+7)=0\\\\ &t=\frac{5}{2}, \frac{-7}{2} \Rightarrow r=2\\\\ & \therefore r=2, a=\frac{1}{2} \end{aligned} $
$ \text { Variance of data set }\left\{\frac{1}{2}, 1,2,4,8\right\} $
$ \therefore \sigma^2=\frac{\sum{\mathrm{X}^2}}{\mathrm{~N}}-\left(\frac{\sum{\mathrm{X}}}{\mathrm{N}}\right)^2 $
$ \begin{aligned} & =\frac{\left(\frac{341}{4}\right)}{5}-\left(\frac{31}{10}\right)^2 \\\\ & =\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100} \\\\ & =\frac{744}{100} \end{aligned} $
$ =\frac{186}{25}=\frac{\mathrm{m}}{\mathrm{n}} \Rightarrow 211=\mathrm{m}+\mathrm{n} $
If the mean of the frequency distribution
| Class : | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency : | 2 | 3 | $x$ | 5 | 4 |
is 28, then its variance is __________.
Explanation:
$ \begin{array}{ll} \text { So, } \frac{2 \times 5+3 \times 15+x \times 25+5 \times 35+4 \times 45}{14+x}=28 \\\\ \Rightarrow \frac{10+45+25 x+175+180}{14+x}=28 \\\\ \Rightarrow 310+25 x=392+28 x \\\\ \Rightarrow 3 x=18 \Rightarrow x=6 \end{array} $
$ \begin{aligned} & \therefore \text { Variance }=\left(\frac{\sum x_i^2 f_i}{\sum f_i}\right)-(\text { mean })^2 \\\\ & =\left(\frac{2 \times 5^2+3 \times 15^2+6 \times 25^2+5 \times 35^2+4 \times 45^2}{20}\right)-(28)^2 \\\\ & =\left(\frac{50+675+3750+6125+8100}{20}\right)-(28)^2 \\\\ & =\left(\frac{18700}{20}\right)-(28)^2 \\\\ & =935-784=151 \end{aligned} $
Let the mean and variance of 8 numbers $x, y, 10,12,6,12,4,8$ be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to _____________.
Explanation:
Now, mean $(\bar{x})=9$
$ \begin{aligned} & \Rightarrow \frac{x+y+52}{8}=9 \\\\ & \Rightarrow x+y=20 \end{aligned} $
Also, variance $=9.25$
$ \begin{aligned} & \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\\\ & \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\\\ & \Rightarrow x^2+y^2-18 \times 20=-142 \\\\ & \Rightarrow x^2+y^2=218 \\\\ & \Rightarrow x^2+(20-x)^2=218 \\\\ & \Rightarrow x^2+400+x^2-40 x=218 \\\\ & \Rightarrow 2 x^2-40 x+182=0 \\\\ & \Rightarrow x=\frac{40 \pm 12}{4} \\\\ & \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13 \\\\ & \text { But } x>y \\\\ & \therefore x=13 \text { and } y=7 \\\\ & \text { So, } 3 x-2 y=39-14=25 \end{aligned} $
Concept :
(a) Mean $=\frac{\Sigma x_i}{n}$
(b) Variance $=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}$
If the mean and variance of the frequency distribution
| $x_i$ | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|
| $f_i$ | 4 | 4 | $\alpha$ | 15 | 8 | $\beta$ | 4 | 5 |
are 9 and 15.08 respectively, then the value of $\alpha^2+\beta^2-\alpha\beta$ is ___________.
Explanation:
$ \therefore $ $\begin{aligned} & \Sigma f_i=40 +\alpha+\beta\end{aligned}$
$\begin{aligned} & \Sigma f_i x_i=360+ 6 \alpha+12 \beta\end{aligned}$
$\begin{aligned} & \Sigma f_i x_i^2=3904 +36 \alpha+144 \beta\end{aligned}$
Given, mean $=9$
$ \begin{aligned} & \Rightarrow \frac{\Sigma f_i x_i}{\Sigma f_i}=9 \\\\ & \Rightarrow \frac{360+6 \alpha+12 \beta}{40+\alpha+\beta}=9 \\\\ & \Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\\\ & \Rightarrow 3 \beta=3 \alpha \\\\ & \Rightarrow \alpha=\beta .........(i) \end{aligned} $
$\begin{aligned} & \text { Variance }=15.08 \\\\ & \Rightarrow \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2=15.08 \\\\ & \Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(9)^2=15.08 \\\\ & \Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}=81+15.08 \quad[\because \alpha=\beta] \\\\ & \Rightarrow 3904+180 \alpha=96.08(40+2 \alpha) \\\\ & \Rightarrow 3904+180 \alpha=3843.2+192.16 \alpha \\\\ & \Rightarrow 60.8=12.16 \alpha \\\\ & \Rightarrow \alpha=5=\beta \\\\ & \therefore \alpha^2+\beta^2-\alpha \beta=25+25-25=25\end{aligned}$
If the variance of the frequency distribution
| $x_i$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| Frequency $f_i$ | 3 | 6 | 16 | $\alpha$ | 9 | 5 | 6 |
is 3, then $\alpha$ is equal to _____________.
Explanation:
$\sigma_{\mathrm{x}}^{2}=\sigma_{\mathrm{d}}^{2}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}^{2}}{\sum \mathrm{f}_{\mathrm{i}}}-\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)^{2}$
$=\frac{150}{45+\alpha}-0=3$
$\Rightarrow 150=135+3 \alpha$
$\Rightarrow 3 \alpha=15 \Rightarrow \alpha=5$
The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then $\mathrm{a+3 b-5}$ is equal to ___________.
Explanation:
$\sum {{x_i} = 7 \times 8 = 56} $
${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} = 16$
${{\sum {x_i^2} } \over 7} - 64 = 16$
$\sum {x_i^2 = 560} $
when 14 is omitted
$\sum {{x_i} = 56 - 14 = 42} $
New mean $ = a = {{\sum {{x_i}} } \over 6} = 7$
$\sum {x_i^2 = 560 - 196 = 364} $
new variance, $b = {{\sum {x_i^2} } \over 6} - {\left( {{{\sum {{x_i}} } \over 6}} \right)^2}$
$ = {{364} \over 6} - 49 = {{35} \over 3}$
$3b = 35$
$a + 3b - 5 = 7 + 35 - 5 = 37$
Let $X=\{11,12,13,....,40,41\}$ and $Y=\{61,62,63,....,90,91\}$ be the two sets of observations. If $\overline x $ and $\overline y $ are their respective means and $\sigma^2$ is the variance of all the observations in $\mathrm{X\cup Y}$, then $\left| {\overline x + \overline y - {\sigma ^2}} \right|$ is equal to ____________.
Explanation:
$x = \{ 11,12,13\,....,40,41\} $
$y = \{ 61,62,63\,....,90,91\} $
$\overline x = {{{{31} \over 2}(11 + 41)} \over {31}} = {1 \over 2} \times 52 = 26$
$\overline y = {{{{31} \over 2}(61 + 91)} \over {31}} = {1 \over 2} \times 152 = 76$
${\sigma ^2} = {{\sum {x_i^2 + \sum {y_i^2} } } \over {62}} - {\left( {{{\sum {x + \sum y } } \over {62}}} \right)^2}$
$ = 705$
Now,
$\left| {\overline x + \overline y - {\sigma ^2}} \right|$
$ = |26 + 76 - 705| = 603$
$ \begin{array}{ccccccc} x_i & 3 & 8 & 11 & 10 & 5 & 4 \\ f_i & 5 & 2 & 3 & 2 & 4 & 4 \end{array} $
Match each entry in List-I to the correct entries in List-II.
| List - I | List - II |
|---|---|
| (P) The mean of the above data is | (1) 2.5 |
| (Q) The median of the above data is | (2) 5 |
| (R) The mean deviation about the mean of the above data is | (3) 6 |
| (S) The mean deviation about the median of the above data is | (4) 2.7 |
| (5) 2.4 |
The correct option is:
The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively. Later on it was found that one of the observations was taken as 50 in the place of 40 . If the wrong entry is replaced by the correct one, then the sum of the squares of all the observations is
162701
163501
162601
161701
The variance of 50 observations is 7 . Suppose that each observation in this data is multiplied by 6 and then 5 is subtracted from it. Then, the variance of that new data is
37
42
247
252