Statistics
If $M$ and $\sigma^2$ represent respectively the mean deviation from the mean and the variance for the data $1,3,5,7$, $11,13,17,19,23$, then $3\left(\sigma^2-M\right)=$
232
112
224
136
If $X$ is a Poisson variate satisfying the condition $3 P(X=2)=P(X=4)$, then $P(X=6)=$
$\frac{162}{5 e^6}$
$\frac{108}{5 e^6}$
$\frac{324}{5 e^6}$
$\frac{648}{5 e^6}$
If the variance of the data $2,3,5,8,12$ is $\sigma^2$ and the mean deviation from the median for this data is $M$, then $\sigma^2-M=$
10.2
5.8
10.6
8.2
Assertion (A) The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.
Reason (R) The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n\left(4 n^2-1\right)}{3}$.
Which of the following alternatives is correct?
The mean and variance of the data $4,5,6,6,7,8, x, y$, where $x< y$ are 6 and $\frac{9}{4}$, respectively. Then, $x^2-2 y$ is equal to
If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is :
The number of values of a $\in$ N such that the variance of 3, 7, 12, a, 43 $-$ a is a natural number is :
Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be ${24 \over 5}$ and ${194 \over 25}$ respectively. If the mean and variance of the first 4 observation are ${7 \over 2}$ and a respectively, then (4a + x5) is equal to:
The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and ${9 \over 4}$ respectively. Then ${x^4} + {y^2}$ is equal to :
The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :
The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to :
Let the mean and the variance of 20 observations $x_{1}, x_{2}, \ldots, x_{20}$ be 15 and 9 , respectively. For $\alpha \in \mathbf{R}$, if the mean of $\left(x_{1}+\alpha\right)^{2},\left(x_{2}+\alpha\right)^{2}, \ldots,\left(x_{20}+\alpha\right)^{2}$ is 178 , then the square of the maximum value of $\alpha$ is equal to ________.
Explanation:
Given $\sum\limits_{{{i = 1} \over {20}}}^{20} {{x_i} = 15 \Rightarrow \sum\limits_{i = 1}^{20} {{x_i} = 300} } $
and $\sum\limits_{{{i = 1} \over {20}}}^{20} {x_i^2 - {{\left( {\overline x } \right)}^2} = 9 \Rightarrow \sum\limits_{i = 1}^{20} {x_i^2 = 4680} } $
Mean $ = {{{{({x_i} + \alpha )}^2} + {{({x_2} + \alpha )}^2}\, + \,.....\, + \,{{({x_{20}} + \alpha )}^2}} \over {20}} = 178$
$ \Rightarrow {{\sum\limits_{i = 1}^{20} {x_i^2 + 2\alpha \sum\limits_{i = 1}^{20} {{x_i} + 20{\alpha ^2}} } } \over {20}} = 178$
$ \Rightarrow 4680 + 600\alpha + 20{\alpha ^2} = 3560$
$ \Rightarrow {\alpha ^2} + 30\alpha + 56 = 0$
$ \Rightarrow {\alpha ^2} + 28\alpha + 2\alpha + 56 = 0$
$ \Rightarrow (\alpha + 28)(\alpha + 2) = 0$
${\alpha _{\max }} = - 2 \Rightarrow \alpha _{\max }^2 = 4.$
Let $x_{1}, x_{2}, x_{3}, \ldots, x_{20}$ be in geometric progression with $x_{1}=3$ and the common ratio $\frac{1}{2}$. A new data is constructed replacing each $x_{i}$ by $\left(x_{i}-i\right)^{2}$. If $\bar{x}$ is the mean of new data, then the greatest integer less than or equal to $\bar{x}$ is ____________.
Explanation:
${x_1},{x_2},{x_3},\,.....,\,{x_{20}}$ are in G.P.
${x_1} = 3,\,r = {1 \over 2}$
$\overline x = {{\sum {x_i^2 - 2{x_i}i + {i^2}} } \over {20}}$
$ = {1 \over {20}}\left[ {12\left( {1 - {1 \over {{2^{40}}}}} \right) - 6\left( {4 - {{11} \over {{2^{18}}}}} \right) + 70 \times 41} \right]$
$\left\{ {\matrix{ {S = 1 + 2\,.\,{1 \over 2} + 3\,.\,{1 \over {{2^2}}}\, + \,....} \cr {{S \over 2} = {1 \over 2} + {2 \over {{2^2}}}\, + \,....} \cr } } \right.$
$\left. {{S \over 2} = 2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}} = 4 - {{11} \over {{2^{18}}}}} \right\}$
$\therefore$ $[\overline x ] = \left[ {{{2858} \over {20}} - \left( {{{12} \over {240}} - {{66} \over {{2^{18}}}}} \right)\,.\,{1 \over {20}}} \right]$
$ = 142$
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _____________.
Explanation:
Given ${{\sum\limits_{i = 1}^{10} {{x_i}} } \over {10}} = 15$ ..... (1)
$ \Rightarrow \sum\limits_{i = 1}^{10} {{x_i} = 150} $
and ${{\sum\limits_{i = 1}^{10} {x_i^2} } \over {10}} - {15^2} = 15$
$ \Rightarrow \sum\limits_{i = 1}^{10} {x_i^2 = 2400} $
Replacing 25 by 15 we get
$\sum\limits_{i = 1}^9 {{x_i} + 25 = 150} $
$ \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 125} $
$\therefore$ Correct mean $ = {{\sum\limits_{i = 1}^9 {{x_i} + 15} } \over {10}} = {{125 + 15} \over {10}} = 14$
Similarly, $\sum\limits_{i = 1}^2 {x_i^2 = 2400 - {{25}^2} = 1775} $
$\therefore$ Correct variance $ = {{\sum\limits_{i = 1}^9 {x_i^2 + {{15}^2}} } \over {10}} - {14^2}$
$ = {{1775 + 225} \over {10}} - {14^2} = 4$
$\therefore$ Correct $S.D = \sqrt 4 = 2$.
The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 \sigma^{2}$ is equal to ___________.
Explanation:
$\mu = {{\sum {{x_i}} } \over {40}} = 30 \Rightarrow \sum {{x_i} = 1200} $
${\sigma ^2} = {{\sum {x_i^2} } \over {40}} - {(30)^2} = 25 \Rightarrow \sum {x_i^2 = 37000} $
After omitting two wrong observations
$\sum {{y_i} = 1200 - 12 - 10 = 1178} $
$\sum {y_i^2 = 37000 - 144 - 100 = 36756} $
Now ${\sigma ^2} = {{\sum {y_i^2} } \over {38}} - {\left( {{{\sum {{y_i}} } \over {38}}} \right)^2}$
$ = {{36756} \over {38}} - {\left( {{{1178} \over {38}}} \right)^2} = - {31^2}$
$ = 38{\sigma ^2} = 36756 - 36518 = 238$
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is _________.
Explanation:
According to given data
${{\sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2}} } \over 7} = 20$
$ \Rightarrow \sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2} = 140} $
So for any xi, ${({x_i} - 62)^2} \le 140$
$ \Rightarrow {x_i} > 50\,\forall i = 1,2,3,\,\,.....\,\,7$
So no student is going to score less than 50.
The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to _____________.
Explanation:
${{\sum {x_i^2} } \over {15}} - {8^2} = 9 \Rightarrow \sum {x_i^2 = 15 \times 73 = 1095} $
Let ${\overline x _c}$ be corrected mean ${\overline x _c}$ = 9
$\sum {x_c^2 = 1095 - 25 + 400 = 1470} $
Correct variance $ = {{1470} \over {15}} - {(9)^2} = 98 - 81 = 17$
If the mean deviation about the mean of the numbers 1, 2, 3, .........., n, where n is odd, is ${{5(n + 1)} \over n}$, then n is equal to ______________.
Explanation:
Mean $ = {{n{{(n + 1)} \over 2}} \over n} = {{n + 1} \over 2}$
M.D. $ = {{2\left( {{{n - 1} \over 2} + {{n - 3} \over 2} + {{n - 5} \over 2} + \,\,\,...\,\,\,0} \right)} \over n} = {{5(n + 1)} \over n}$
$ \Rightarrow ((n - 1) + (n - 3) + (n - 5) + \,\,...\,\,0) = 5(n + 1)$
$ \Rightarrow \left( {{{n + 1} \over 4}} \right)\,.\,(n - 1) = 5(n + 1)$
So, $n = 21$
Statement I The range of the ungrouped data does not change even if certain intermediate observations are removed
Statement II The value of the mean deviation of an ungrouped data about the median is always less than or equal to the value of the mean deviation computed about any other measure of central tendency
Statement III For a grouped data, range is approximated as the difference between the lower limit of the largest class and the upper limit of the smallest class
Statements I and II are true but Statement III is false
Statements II and III are true but Statement I is false
Statement III and I are true but Statement II is false
Statements I, II and III are true
If 10 is the mean deviation of ' $n$ ' observations $x_1, x_2, x_3, \ldots, x_n$, then the mean deviation of the observations $\frac{2 x_1+5}{3}, \frac{2 x_2+5}{3}, \frac{2 x_3+5}{3}, \ldots . \frac{2 x_n+5}{3}$ is
$25 / 3$
$40 / 9$
$20 / 3$
15
There are $n$ observations and all of them are negative numbers. The ascending order of these observations is $x_1, x_2, \ldots . x_n$. If the signs of the first term and last term in that order are changed, then the range of the data is
$\left|x_1\right|-\left|x_n\right|$
$\left|x_n-x_1\right|$
$\left|x_1\right|-x_2$
$\left|x_1\right|-\left|x_2\right|$
The mean deviation from the mean for the observations $1,3,5,7,11,13,17,19,23$ is
6
$11 \frac{4}{9}$
11
$6 \frac{2}{9}$
The mean deviation from the mean of the discrete data $1,3,4,7,11,18,29,47,78$ is
22
24
$\frac{176}{9}$
$\frac{182}{9}$
If $\bar{x}$ is the mean of $n$ observations $x_1, x_2, \ldots ., x_n$ then the mean of the absolute deviations of these observations from $\bar{x}$ is
the variance of the data
the mean proportion of the data
the standard deviation of the data
the mean deviation of the data
If the mean deviation of the data $1,1+d 1+2 d, \ldots, 1+100 d,(d>0)$ from their mean is 255, then '$d$' is equal to
If the mean of the data $p, 6,6,7,8,11,15,16$, is 3 times $p$, then the mean deviation of the data from its mean is
The mean deviation about the mean for the following data.
$5,6,7,8,6,9,13,12,15 \text { is }$
The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are
$\matrix{ {x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \cr {f:} & 4 & 4 & \alpha & \beta \cr } $
be 6 and 6.8 respectively. If x3 is changed from 8 to 7, then the mean for the new data will be :
respectively, then (a $-$ b)2 is equal to :
Explanation:
Tn = (3n + 4) (2n + 6) = 2(3n + 4) (n + 3)
= 2(3n2 + 13n + 12) = 6n2 + 26n + 24
S10 = $\sum\limits_{n = 1}^{10} {{T_n}} = 6\sum\limits_{n = 1}^{10} {{n^2}} + 26\sum\limits_{n = 1}^{10} n + 24\sum\limits_{n = 1}^{10} 1 $
$ = {{6(10 \times 11 \times 21)} \over 6} + 26 \times {{10 \times 11} \over 2} + 24 \times 10$
$ = 10 \times 11(21 + 13) + 240$
= 3980
Mean $ = {{{S_{10}}} \over {10}} = {{3980} \over {10}} = 398$
Explanation:
n1 = no. of boys
${\overline x _b}$ = 12
n2 = no. of girls
$\sigma _g^2$ = 2
${\overline x _g}$ = ${{50 \times 15 - 12 \times {\sigma _b}} \over {30}} = {{750 - 12 \times 20} \over {30}} = 17 = \mu $
variance of combined series
$\sigma _{}^2 = {{{n_1}\sigma _b^2 + {n_2}\sigma _g^2} \over {{n_1} + {n_2}}} + {{{n_1}.\,{n_2}} \over {{{({n_1} + {n_2})}^2}}}{\left( {{{\overline x }_b} - {{\overline x }_g}} \right)^2}$
$\sigma _{}^2 = {{20 \times 2 + 30 \times 2} \over {20 + 30}} + {{20 \times 30} \over {{{(20 + 30)}^2}}}{(12 - 17)^2}$
$\sigma$2 = 8
$\Rightarrow$ $\mu$ + $\sigma$2 = 17 + 8 = 25
Explanation:
Explanation:
Var(x) = $10 = {{{3^2} + {7^2} + {x^2} + {y^2}} \over 4} - 25$
$140 = 49 + 9 + {x^2} + {y^2}$
${x^2} + {y^2} = 82$
x + y = 10
$\Rightarrow$ (x, y) = (9, 1)
Four numbers are 21, 9, 10, 8
Mean = ${{48} \over 4}$ = 12
| Class : | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|
| Frequency : | $\alpha $ | 110 | 54 | 30 | $\beta $ |
If the sum of all frequencies is 584 and median is 45, then | $\alpha$ $-$ $\beta$ | is equal to _______________.
Explanation:
$\Rightarrow$ $\alpha$ + $\beta$ = 390
Now, median is at ${{584} \over 2}$ = 292th
$\because$ Median = 45 (lies in class 40 - 50)
$\Rightarrow$ $\alpha$ + 110 + 54 + 15 = 292
$\Rightarrow$ $\alpha$ = 113, $\beta$ = 277
$\Rightarrow$ | $\alpha$ $-$ $\beta$ | = 164
| Class : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
|---|---|---|---|---|---|
| Frequency : | $a $ | $b$ | 12 | 9 | 5 |
If mean = ${{309} \over {22}}$ and median = 14, then the value (a $-$ b)2 is equal to _____________.
Explanation:
| Class | Frequency | ${x_i}$ | ${f_i}{x_i}$ |
|---|---|---|---|
| 0-6 | a | 3 | 3a |
| 6-12 | b | 9 | 9b |
| 12-18 | 12 | 15 | 180 |
| 18-24 | 9 | 21 | 189 |
| 24-30 | 5 | 27 | 135 |
| $N = (26 + a + b)$ | $(504 + 3a + 9b)$ |
Mean = ${{3a + 9b + 180 + 189 + 135} \over {a + b + 26}} = {{309} \over {22}}$
$ \Rightarrow 66a + 198b + 11088 = 309a + 309b + 8034$
$ \Rightarrow 243a + 111b = 3054$
$ \Rightarrow 81a + 37b = 1018$ $\to$ (1)
Now, Median $ = 12 + {{{{a + b + c} \over 2} - (a + b)} \over {12}} \times 6 = 14$
$ \Rightarrow {{13} \over 2} - \left( {{{a + b} \over 4}} \right) = 2$
$ \Rightarrow {{a + b} \over 4} = {9 \over 2}$
$ \Rightarrow a + b = 18$ $\to$ (2)
From equation (1) $ (2)
a = 8, b = 10
$\therefore$ ${(a - b)^2} = {(8 - 10)^2}$ = 4
Explanation:
Here, Mean = 40 of 25 teachers
$\therefore$ 40 = ${{\sum x } \over {25}}$
$ \Rightarrow $ $\sum x $ = 40 $ \times $ 25 = 1000
After retireing of a 60 year old teacher, total age of 24 teachers,
x1 + x2 + . . . . . .x24 = 1000 $-$ 60 = 940
Now a new teacher of age A year is appointed.
$\therefore$ Now total age of this 25 teachers
x1 + x2 + x3 + . . . . . + x25 = 940 + A
$\therefore$ Mean age = ${{940 + A} \over {25}}$
According to question,
${{940 + A} \over {25}}$ = 39
$ \Rightarrow $ A = 35
Explanation:
and last n observations are y1, y2 ....................., yn
Now, ${{\sum {{x_i}} } \over {2n}} = 6$, ${{\sum {{y_i}} } \over n} = 3$
$ \Rightarrow \sum {{x_i}} = 12n,\sum {{y_i}} = 3n$
$ \therefore $ ${{\sum {{x_i}} + \sum {{y_i}} } \over {3n}} = {{15n} \over {3n}} = 5$
Now, ${{\sum {x_i^2} + \sum {y_i^2} } \over {3n}} - {5^2} = 4$
$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n = 87n$
Now, mean is ${{\sum {({x_i} + 1) + \sum {({y_i} - 1)} } } \over {3n}} = {{15n + 2n - n} \over {3n}} = {{16} \over 3}$
Now, variance is ${{{{\sum {{{({x_i} + 1)}^2} + \sum {({y_i} - 1)} } }^2}} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$
$ = {{\sum {x_i^2 + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n} } \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$
$ = {{87n + 2(9n) + 3n} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$
= $29 + 6 + 1 - {\left( {{{16} \over 3}} \right)^2}$
$ = {{324 - 256} \over 9} = {{68} \over 9} = k$
$ \Rightarrow $ 9k = 68
Therefore, the correct answer is 68.
| Size | Mean | Variance | |
|---|---|---|---|
| Observation I | 10 | 2 | 2 |
| Observation II | n | 3 | 1 |
If the variance of the combined set of these two observations is ${{17} \over 9}$, then the value of n is equal to ___________.
Explanation:
${{\sum {{x_i^2}} } \over {10}} - {(2)^2} = 2 \Rightarrow \sum {x_i^2} = 60$
For group - 2 : ${{\sum {{y_i}} } \over n} = 3 \Rightarrow \sum {{y_i}} = 3n$
${{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \sum {y_i^2} = 10n$
Now, combined variance
${\sigma ^2} = {{\sum {\left( {x_i^2 + y_i^2} \right)} } \over {10 + n}} - {\left( {{{\sum {\left( {{x_i} + {y_i}} \right)} } \over {10 + n}}} \right)^2}$
$ \Rightarrow {{17} \over 9} = {{60 + 10n} \over {10 + n}} - {{{{(20 + 3n)}^2}} \over {{{(10 + n)}^2}}}$
$ \Rightarrow $ 17 (n2 + 20n + 100) = 9(n2 + 40n + 200)
$ \Rightarrow $ 8n2 $-$ 20n $-$ 100 = 0
$ \Rightarrow $ 2n2 $-$ 5n $-$ 25 = 0 $ \Rightarrow $ n = 5
that $\sum\limits_{i = 1}^{18} {({X_i} - } \alpha ) = 36$ and $\sum\limits_{i = 1}^{18} {({X_i} - } \beta {)^2} = 90$, where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of | $\alpha$ $-$ $\beta$ | is ____________.
Explanation:
$ \Rightarrow \sum {{x_i} - 18\alpha = 36} $
$ \Rightarrow \sum {{x_i} = 18(\alpha + 2)} $ .... (1)
Also, $\sum\limits_{i = 1}^{18} {{{({x_1} - \beta )}^2} = 90} $
$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} - 2\beta \sum {{x_i} = 90} } $
$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} + 2\beta \times 18(\alpha + 2) = 90} $ (using equation (1))
$ \Rightarrow \sum {x_i^2 = 90} - 18{\beta ^2} + 36\beta (\alpha + 2)$
Given, ${\sigma ^2} = 1 \Rightarrow {1 \over {18}}{\sum {x_i^2 - \left( {{{\sum {{x_i}} } \over {18}}} \right)} ^2} = 1$
$ = {1 \over {18}}(90 - 18{\beta ^2} + 36\alpha \beta + 72\beta ) - {\left( {{{18(\alpha + 2)} \over {18}}} \right)^2} = 1$
$ \Rightarrow 90 - 18{\beta ^2} + 36\alpha \beta + 72\beta - 18{(\alpha + 2)^2} = 18$
$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {(\alpha + 2)^2} = 1$
$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {\alpha ^2} - 4 - 4\alpha = 1$
$ \Rightarrow {\alpha ^2} - {\beta ^2} + 2\alpha \beta + 4\beta - 4\alpha = 0$
$ \Rightarrow (\alpha - \beta )(\alpha - \beta + 4) = 0$
$ \Rightarrow \alpha - \beta = - 4$
$ \therefore $ $|\alpha - \beta |\, = 4$ $(\alpha \ne \beta )$
Explanation:
${\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} < 10$
$(90 + {k^2})10 - (81 + {k^2} + 8k) < 1000$
$90 + 10{k^2} - {k^2} - 18k - 81 < 1000$
$9{k^2} - 18k + 9 < 1000$
${(k - 1)^2} < {{1000} \over 9} \Rightarrow k - 1 < {{10\sqrt {10} } \over 3}$
$k < {{10\sqrt {10} } \over 3} + 1$
k $ \le $ 11
Maximum integral value of k = 11.
If the mean of a data x is 10 and if all the observations are multiplied by 2, then the mean of new data is
The mean deviation from the mean of the set of observation $-1,0,4$ is