Statistics
83 JEE Maths previous year questions on Statistics — options free on every question; 8 include the answer & explanation free, the rest unlock with PYQ Pass.
If the mean deviation about the median of the numbers \(k, 2 k, 3 k, \ldots, 1000 k\) is \(500\), then \(k^2\) is equal to:
[JEE Main 2026, 22 Jan (Shift 2)]
\(4\)
Here number of terms are even therefore
median \(=\frac{500k+501k}{2}=\frac{1001\text{k}}{2}={\text{X}}_{\text{M}}\)
mean deviation about median \(=\frac{\sum \left|{\text{X}}_{\text{i}}−{\text{X}}_{\text{M}}\right|}{\text{n}}\)
\(=\frac{2\left(\frac{\text{k}}{2}+\frac{3\text{k}}{2}+\frac{5\text{k}}{2}+\ldots 500\text{ terms }\right)}{1000}\)
\(=\frac{2\times \frac{k}{2}\left(1+3+5+.....500\mathrm{terms}\right)}{1000}\\ =\frac{2\times \frac{k}{2}\left(\frac{500}{2}\left\{2+\left(500-1\right)\times 2\right\}\right)}{1000}\)
\(=\frac{2 \cdot \frac{k}{2}(500)^2}{1000}=\frac{500 k}{2}=500(\) given \()\)
\(\therefore k=2\)
Hence, \(k^2=4\)
Let the mean and variance of 8 numbers \(-10,-7,-1,x,y,9,2,16\) be \(\frac{7}{2}\) and \(\frac{293}{4}\), respectively. Then the mean of 4 numbers \(x,y,x+y+1,|x-y|\) is:
[JEE Main 2026, 23 Jan (Shift 2)]
\(11\)
Given: Mean
\(\mu =\frac{-18+x+y+2+9+16}{8}=\frac{7}{2}\\ \Rightarrow \frac{x+y+9}{8}=\frac{7}{2}\\ \Rightarrow x+y+9=28\\ \Rightarrow x+y=19....\left(i\right)\)
And Variance
\({\sigma }^{2}=\frac{\sum {x}_{i}^{2}}{8}-{\left(\mu \right)}^{2}=\frac{293}{4}\\ \Rightarrow \frac{{10}^{2}+{7}^{2}+{1}^{2}+{x}^{2}+{y}^{2}+{2}^{2}+{9}^{2}+{16}^{2}}{8}-{\left(\frac{7}{2}\right)}^{2}=\frac{293}{4}\)
\(\Rightarrow \frac{293}{4}+\frac{49}{4}=\frac{{x}^{2}+{y}^{2}+491}{8}\)
\(\Rightarrow \frac{171}{2}=\frac{{x}^{2}+{y}^{2}+491}{8}\)
\(\Rightarrow {x}^{2}+{y}^{2}=193...\left(ii\right)\)
After solving equation \((i)\) and \((ii)\), We get:
\(x=7\) and \(y=12\) or \(x=12\) and \(y=7\)
Hence numbers are \(7,12,20,5\)
Mean\(=\frac{20+12+7+5}{4}=\frac{44}{4}=11\)
The mean and variance of a data of \(10\) observations are \(10\) and \(2\) , respectively. If an observations \(\alpha\) in this data is replaced by \(\beta\), then the means and variance becomes \(10.1\) and \(1.99\), respectively. Then \(\alpha+\beta\) equals
\(20\)
Let, in first case, \(10\) numbers are \(x_1, x_2, \ldots \ldots x_9, \alpha\)
Then, \(\frac{\sum _{i=1}^{9}{x}_{i}+\alpha }{10}=10\)
\(\Rightarrow \alpha +\sum _{i=1}^{9}{x}_{i}=100\Rightarrow \sum _{i=1}^{9}{x}_{i}=100−\alpha\)
And \({\sigma }^{2}=\left(\frac{\sum {x}_{i}^{2}+{\alpha }^{2}}{n}\right)−{\left(\frac{\sum {x}_{i}+\alpha }{n}\right)}^{2}\)
\(\Rightarrow 2=\left(\frac{\sum {x}_{i}^{2}+{\alpha }^{2}}{10}\right)−{\left(10\right)}^{2}\)
\(\Rightarrow \frac{\sum {x}_{i}^{2}+{\alpha }^{2}}{n}=102\)
\(\Rightarrow {x}_{1}^{2}+{x}_{2}^{2}+\ldots \ldots {x}_{9}^{2}+{\alpha }^{2}=1020\)
\(\Rightarrow \sum {x}_{i}^{2}=1020−{\alpha }^{2}\)
In second case, let number are
\({x}_{1},{x}_{2},\ldots \ldots {x}_{9},\beta\)
Now \(\frac{\sum_{i=1}^9 x_i+\beta}{10}=\frac{100-\alpha+\beta}{10}\)
\(\Rightarrow 10.1=\frac{100-\alpha+\beta}{10}\)
\(\Rightarrow \alpha-\beta=-1\)
Now, \({\sigma }^{2}=\left(\frac{\sum {x}_{i}^{2}+{\beta }^{2}}{n}\right)−{\left(\frac{\sum {x}_{i}+\beta }{n}\right)}^{2}\)
\(\Rightarrow \frac{\sum {x}_{i}^{2}+{\beta }^{2}}{10}−{(10.1)}^{2}=1.99\)
\(\Rightarrow 1.99=\frac{1020-{\alpha }^{2}+{\beta }^{2}}{10}-102.01\\ \Rightarrow 1020-{\alpha }^{2}+{\beta }^{2}=1040\)
\(\Rightarrow {\beta }^{2}−{\alpha }^{2}=20\)
\(\Rightarrow \alpha^2-\beta^2=-20\)
\(\Rightarrow (\alpha-\beta)(\alpha+\beta)=-20 \)
\(\Rightarrow (-1)(\alpha+\beta)=-20\)
Hence, \(\alpha+\beta=20\)
For \(10\) observations \({x}_{1},{x}_{2},\ldots ,{x}_{10}\), if \(\sum _{i=1}^{10}{\left({x}_{i}+2\right)}^{2}=180\) and \(\sum _{i=1}^{10}{\left({x}_{i}-1\right)}^{2}=90\), then their standard deviation is:
[JEE Main 2026, 4 Apr (Shift 2)]
\(3\)
\(\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}+2\right)^2=180 \)
\( \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}+\sum_{\mathrm{i}=1}^{10} 4=180 \)
\( \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=180-40=140 \ldots\ (1)\)
\( \text { Also } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-1\right)^2=90 \)
\( \sum_{i=1}^{10} x_i^2-2 \sum_{i=1}^{10} x_i+\sum_{i=1}^{10} 1=90 \)
\( \sum_{i=1}^{10} x_i^2-2 \sum_{i=1}^{10} x_i=90-10=80 \ldots\ (2)\)
From (1) and (2) we get:
\( \sum_{i=1}^{10} x_i^2=100 \text { and } \sum_{i=1}^{10} x_i=10 \)
\( \sigma^2=\frac{\sum_{i=1}^{10} x_i^2}{N}-\left(\frac{\sum_{i=1}^{10} x_i}{N}\right)^2\)
\( =\frac{100}{10}-\left(\frac{10}{10}\right)^2\)
\( \sigma^2=10-1=9 \)
\( \Rightarrow \sigma=3\)
Let the mean and the variance of seven observations \(2,4, \alpha, 8, \beta,12,14, \alpha<\beta\), be \(8\) and \(16\) respectively. Then the quadratic equation whose roots are \(3 \alpha+2\) and \(2 \beta+1\) is:
[JEE Main 2026, 6 Apr (Shift 2)]
\({x}^{2}-41x+420=0\)
Mean \(\bar{x}=\frac{\sum x_i}{n}\)
\(8=\frac{2+4+\alpha+8+\beta+12+14}{7}\)
\(56=40+\alpha+\beta\)
\(\alpha+\beta=16\)
and \(\sigma^2=\frac{\sum x_i^2}{n}-(\bar{x})^2\)
\(16=\frac{4+16+\alpha^2+64+\beta^2+144+196}{7}-64\)
\(560=424+\alpha^2+\beta^2\)
\(\alpha^2+\beta^2=136\)
Since \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\)
\(256=136+2 \alpha \beta\)
\(\alpha \beta=60\)
\(\alpha\) and \(\beta\) are the roots of the quadratic equation \(t^2-(\alpha+\beta) t+\alpha \beta=0\)
\(t^2-16 t+60=0\)
\((t-6)(t-10)=0\)
\(\alpha=6\) and \(\beta=10\)
Now, \(3 \alpha+2=20\) and \(2 \beta+1=21\)
Sum of roots \(=20+21=41\)
Product of roots \(=420\)
Required equation is
\(x^2-41 x+420=0\)
If the variance of the frequency distribution
\(\begin{matrix}x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2 & 1 & 1 & 1 & 1 & 1\end{matrix}\)
is 160 , then the value of \(c\in N\) is
[JEE Main 2024, 09 Apr (Shift 2)]
7
\(\begin{matrix}x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2 & 1 & 1 & 1 & 1 & 1\end{matrix}\)
Mean \(\overset{¯}{x}=\frac{(2+2+3+4+5+6)C}{7}=\frac{22C}{7}\)
variance \({\sigma }^{2}=\frac{{c}^{2}(2+{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}+{6}^{2})}{7}-{\left(\frac{22c}{7}\right)}^{2}\)
\(=\frac{92{c}^{2}}{7}-{c}^{2}\times \frac{484}{49}\)
\(=\frac{(644-484){c}^{2}}{49}=\frac{160{c}^{2}}{49}\)
\(160=\frac{160\times {c}^{2}}{49}\Rightarrow c=7\)
If the variance of the frequency distribution
\(\begin{matrix}x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2 & 1 & 1 & 1 & 1 & 1\end{matrix}\)
is 160 , then the value of \(c\in N\) is
[JEE Main 2024, 09 Apr (Shift 2)]
7
\(\begin{matrix}x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2 & 1 & 1 & 1 & 1 & 1\end{matrix}\)
Mean \(\overset{¯}{x}=\frac{(2+2+3+4+5+6)C}{7}=\frac{22C}{7}\)
variance \({\sigma }^{2}=\frac{{c}^{2}(2+{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}+{6}^{2})}{7}-{\left(\frac{22c}{7}\right)}^{2}\)
\(=\frac{92{c}^{2}}{7}-{c}^{2}\times \frac{484}{49}\)
\(=\frac{(644-484){c}^{2}}{49}=\frac{160{c}^{2}}{49}\)
\(160=\frac{160\times {c}^{2}}{49}\Rightarrow c=7\)
The mean of 10 observations is 5.5. If each observation is multiplied by 4 and subtracted from 44, then what is the new mean?
22
New mean = 44 – 4 × 5.5 = 44 – 22 = 22
For a statistical data \({x}_{1},{x}_{2},\ldots ,{x}_{10}\) of 10 values, a student obtained the mean as \(5.5\) and \(\sum _{i=1}^{10}{x}_{i}^{2}=371.\) He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is
[JEE Main 2025, 24 Jan (Shift 1)]
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10 is the mean of a set of 7 observations and 5 is the mean of a set of 3 observations. The mean of the combined set is given by
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Let the Mean and Variance of five observations \({\mathrm{x}}_{1}=1,{\mathrm{x}}_{2}=3,{\mathrm{x}}_{3}=\mathrm{a},{\mathrm{x}}_{4}=7\) and \({\mathrm{x}}_{5}=\mathrm{b},\mathrm{a}>\mathrm{b}\), be \(5\) and \(10\) respectively. Then the Variance of the observations \(\mathrm{n}+{\mathrm{x}}_{\mathrm{n}},\mathrm{n}=1,2,\ldots \ldots ..5\) is
[JEE Main 2025, 3 Apr (Shift 2)]
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Let \(a_1, a_2, \ldots a_{10}\) be 10 observations such that \(\sum_{k=1}^{10} a_k=50\) and \(\sum _{\forall k [JEE Main 2024, 27 Jan (Shift 1)]
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Let \(a_1, a_2, \ldots a_{10}\) be 10 observations such that \(\sum_{k=1}^{10} a_k=50\) and \(\sum _{\forall k [JEE Main 2024, 27 Jan (Shift 1)]
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Marks obtains by all the students of class \(12\) are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be \(14\) with median class interval \(12-18\) and median class frequency \(12.\) If the number of students whose marks are less than \(12\) is \(18,\) then the total number of students is
[JEE Main 2025, 23 Jan (Shift 1)]
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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 \(\sigma^2-M\) is:
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A data consists of \(20\) observations \({x}_{1},{x}_{2},\ldots ,{x}_{20}\). If \(\sum _{i=1}^{20}{\left({x}_{i}+5\right)}^{2}=2500\) and \(\sum _{i=1}^{20}{\left({x}_{i}-5\right)}^{2}=100\), then the ratio of mean to standard deviation of this data is:
[JEE Main 2026, 6 Apr (Shift 1)]
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Marks obtains by all the students of class \(12\) are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be \(14\) with median class interval \(12-18\) and median class frequency \(12.\) If the number of students whose marks are less than \(12\) is \(18,\) then the total number of students is
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For a distribution of 10 observations, \(\sum_{i=1}^{10} x_i=55\) and \(\sum_{i=1}^{10} x_i^2=328\) If the observations 4 and 5 are replaced by 6 and 8 respectively, then the new variance is
[JEE Main 2025]
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Let \({\mathrm{x}}_{1},{\mathrm{x}}_{2},\ldots \ldots {\mathrm{x}}_{10}\) be ten observations such that \(\sum _{i=1}^{10}\left({x}_{i}-2\right)=30,\sum _{i=1}^{10}{\left({x}_{i}-\beta \right)}^{2}=98,\beta >2\) and their variance is \(\frac{4}{5}\). If \(\mu\) and \({\sigma }^{2}\) are respectively the mean and the variance of \(2\left({x}_{1}-1\right)+4\beta ,2\left({x}_{2}-1\right)+4\beta ,\)\(\ldots ..,2\left({x}_{10}-1\right)+4\beta ,\) then \(\frac{\beta \mu }{{\sigma }^{2}}\) is equal to :
[JEE Main 2025, 29 Jan (Shift 1)]
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A variable \(X\) takes values \(0,0,2,6,12,20, \ldots \mathrm{n}(\mathrm{n}-1)\) with frequencies \({ }^{\mathrm{n}} \mathrm{C}_0,{ }^{\mathrm{n}} \mathrm{C}_1,{ }^{\mathrm{n}} \mathrm{C}_2,{ }^{\mathrm{n}} \mathrm{C}_3,{ }^{\mathrm{n}} \mathrm{C}_4,{ }^{\mathrm{n}} \mathrm{C}_5, \ldots,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}\), respectively. If the mean of this data is \(60\), then its median is:
[JEE Main 2026, 5 Apr (Shift 2)]
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Let the mean and the standard deviation of the observation \(2,3,3,4,5,7,\mathrm{a},\mathrm{b}\) be \(4\) and \(\sqrt{2}\) respectively. Then the mean deviation about the mode of these observations is :
[JEE Main 2025, 4 Apr (Shift 2)]
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Marks obtained by all the students of class \(12^{\text {th }}\) are in a frequency distribution with classes of equal width. Let the median of the group data be 14 with median class interval 12-18 and the median class frequency is 12 . If the number of students who secures marks below 12 is 18 then the total number of students is
[JEE Main 2025]
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The mean and variance of \(10\) observations are \(9\) and \(34.2\), respectively. If \(8\) of these observations are \(2, 3, 5, 10, 11, 13, 15, 21\), then the mean deviation about the median of all the \(10\) observations is
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For a distribution of 10 observations, \(\sum_{i=1}^{10} x_i=55\) and \(\sum_{i=1}^{10} x_i^2=328\) If the observations 4 and 5 are replaced by 6 and 8 respectively, then the new variance is
[JEE Main 2025]
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The mean and variance of \(10\) observations are \(9\) and \(34.2\), respectively. If \(8\) of these observations are \(2, 3, 5, 10, 11, 13, 15, 21\), then the mean deviation about the median of all the \(10\) observations is
[JEE Main 2026, 28 Jan (Shift 1)]
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Let \({\mathrm{x}}_{1},{\mathrm{x}}_{2},\ldots \ldots {\mathrm{x}}_{10}\) be ten observations such that \(\sum _{i=1}^{10}\left({x}_{i}-2\right)=30,\sum _{i=1}^{10}{\left({x}_{i}-\beta \right)}^{2}=98,\beta >2\) and their variance is \(\frac{4}{5}\). If \(\mu\) and \({\sigma }^{2}\) are respectively the mean and the variance of \(2\left({x}_{1}-1\right)+4\beta ,2\left({x}_{2}-1\right)+4\beta ,\)\(\ldots ..,2\left({x}_{10}-1\right)+4\beta ,\) then \(\frac{\beta \mu }{{\sigma }^{2}}\) is equal to :
[JEE Main 2025, 29 Jan (Shift 1)]
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Let the mean and the standard deviation of the observation \(2,3,3,4,5,7,\mathrm{a},\mathrm{b}\) be \(4\) and \(\sqrt{2}\) respectively. Then the mean deviation about the mode of these observations is :
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If the mean and the variance of \(6,4,a,8,b,12,10,13\), are \(9\) and \(9.25\) respectively, then \(\mathrm{a}+\mathrm{b}+\mathrm{ab}\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
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For a statistical data \({x}_{1},{x}_{2},\ldots ,{x}_{10}\)of 10 values, a student obtained the mean as \(5.5\) and \(\sum _{i=1}^{10}{x}_{i}^{2}=371.\) He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is
[JEE Main 2025, 24 Jan (Shift 1)]
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The mean and standard deviation of \(100\) observations are \(40\)and \(5.1\), respectively, By mistake one observation is taken as \(50\) instead of \(40\). If the correct mean and the correct standard deviation are \(\mu\) and \(\sigma\) respectively, then \(10(\mu +\sigma )\) is equal to
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If mean of \(n\) item is \(\overset{¯}{x}\). If each \({r}^{\mathrm{th}}\) item is increased by \(2r\). Then new mean will be
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Marks obtained by all the students of class \(12^{\text {th }}\) are in a frequency distribution with classes of equal width. Let the median of the group data be 14 with median class interval 12-18 and the median class frequency is 12 . If the number of students who secures marks below 12 is 18 then the total number of students is
[JEE Main 2025]
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Let \(X=\{x \in N: 1 \leq x \leq 19\}\) and for some \(a, b \in R, Y=\{a x+b: x \in X\}\). If the mean and variance of the elements of \(Y\) are \(30\) and \(750\), respectively, then the sum of all possible values of \(b\) is
[JEE Main 2026, 24 Jan (Shift 2)]
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The mean and variance of \(n\) observations are \(8\) and \(16\), respectively. If the sum of the first \((n – 1)\) observations is \(48\) and the sum of squares of the first \((n – 1)\) observations is \(496,\) then the value of \(n\) is:
[JEE Main 2026, 2 Apr (Shift 2)]
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A set of four observations has mean \(1\) and variance \(13\). Another set of six observations has mean \(2\) and variance \(1\). Then, the variance of all these \(10\) observations is equal to:
[JEE Main 2026, 8 Apr (Shift 2)]
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Let the Mean and Variance of five observations \({\mathrm{x}}_{1}=1,{\mathrm{x}}_{2}=3,{\mathrm{x}}_{3}=\mathrm{a},{\mathrm{x}}_{4}=7\) and \({\mathrm{x}}_{5}=\mathrm{b},\mathrm{a}>\mathrm{b}\), be \(5\) and \(10\) respectively. Then the Variance of the observations \(\mathrm{n}+{\mathrm{x}}_{\mathrm{n}},\mathrm{n}=1,2,\ldots \ldots ..5\) is
[JEE Main 2025, 3 Apr (Shift 2)]
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10 is the mean of a set of 7 observations and 5 is the mean of a set of 3 observations. The mean of the combined set is given by
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If the mean deviation about the median of the numbers \(k, 2 k, 3 k, \ldots, 1000 k\) is \(500\), then \(k^2\) is equal to:
[JEE Main 2026, 22 Jan (Shift 2)]
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Suppose that the mean and median of the non-negative numbers \(21,8,17,a,51,103,b,13,67,(a>b),\) are \(40\) and \(21,\) respectively. If the mean deviation about the median is \(26,\) then \(2a\) is equal to:
[JEE Main 2026, 4 Apr (Shift 1)]
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Let \(X=\{x \in N: 1 \leq x \leq 19\}\) and for some \(a, b \in R, Y=\{a x+b: x \in X\}\). If the mean and variance of the elements of \(Y\) are \(30\) and \(750\), respectively, then the sum of all possible values of \(b\) is
[JEE Main 2026, 24 Jan (Shift 2)]
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The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is
[JEE Main 2024, 06 Apr (Shift 1)]
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The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is
[JEE Main 2024, 06 Apr (Shift 1)]
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If the mean and the variance of \(6,4,a,8,b,12,10,13\), are \(9\) and \(9.25\) respectively, then \(\mathrm{a}+\mathrm{b}+\mathrm{ab}\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
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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 \(\sigma^2-M\) is:
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Let the mean and variance of 8 numbers \(-10,-7,-1,x,y,9,2,16\) be \(\frac{7}{2}\) and \(\frac{293}{4}\), respectively. Then the mean of 4 numbers \(x,y,x+y+1,|x-y|\) is:
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Let \(X=\{11,12,13, \ldots . ., 40,41\}\) and \(Y=\{61,62,63, \ldots \ldots\), \(90,91\}\) be the two sets of observations. If \(\bar{x}\) and \(\bar{y}\) are their respective means and \(\sigma^2\) is the variance of all the observations in \(X \cup Y\), then \(\left|\bar{x}+\bar{y}-\sigma^2\right|\) is equal to__________
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Find the mean deviation about the mean for the data
4, 7, 8, 9, 10, 12, 13, 17
[JEE Main 2023]
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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 \(A\) and adding 2 to each element of \(B\). Then the sum of the mean and variance of the elements of \(C\) is
[JEE Main 2023, 11 Apr (Shift 1)]
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Let in a series of \( 2 \mathrm{n} \) observations, half of them are equal to \(a\) and remaining half are equal to \(-a\). Also by adding \( a \) constant \( b \) in each of these observations, the mean and standard deviation of new set become 5 and 20 , respectively. Then the value of \( a^{2}+b^{2} \) is equal to:
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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..........
[JEE Main 2023, 13 Apr (Shift 2)]
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Consider three observations \(a, b\) and \(c\) such that \(b=a+c\). If the standard deviation of \(a+2, b+2, c+2\) is \(d\), then which of the following is true?
[JEE Main 2021, 16 Mar (Shift 1)]
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If the mean and variance of the following data \(6,10,7,13,a,12,b,12\) are\(9\)and \(\frac{37}{4}\) respectively, then \({(a-b)}^{2}\) is equal to
[JEE Main 2021, 27 Jul (Shift 1)]
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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 \(A\) and adding \(2\) to each element of \(B\). Then the sum of the mean and variance of the elements of \(C\) is
[JEE Main 2023, 11 Apr (Shift 1)]
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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
[JEE Main 2023, 1 Feb (Shift 1)]
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The mean of 10 numbers \(7 \times \ 8,10 \times \ 10,13 \times \ 12,16 \times \ 14 ,\ ....\ \) is
[JEE Main 2021, 31 Aug (Shift 1)]
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The first of the two samples has 100 items with mean 15 and \(\mathrm{S}.\mathrm{D}.3\). If the whole group has 250 items with mean 15.6 and S.D\(=\sqrt{13.44}\) then S.D. of the second group is
[JEE Main 2021, 25 Jul (Shift 2)]
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Let \(S\) be the set of all values of \(a_1\) for which the mean deviation about the mean of 100 consecutive positive integers \(a_1, a_2, a_3, \ldots ., a_{100}\) is 25 . Then \(S\) is
[JEE Main 2023, 30 Jan (Shift 2)]
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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 \(m\) and \(n\) are co-prime, then \(m+n\) is equal to
[JEE Main 2023, 8 Apr (Shift 2)]
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If the mean and variance of the following data :
6,10,7,13, a, 12, b, 12 are 9 and \(\frac{37}{4}\) respectively, then \((a-b{)}^{2}\) is equal to :
[JEE Main 2021, 27 Jul (Shift 1)]
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Let in a series of \(2 n\) observations, half of them are equal to \(a\) and remaining half are equal to \(-a\). Also by adding a constant \(b\) in each of these observations, the mean and standard deviation of new set become 5 and 20 , respectively. Then the value of \(a^2+b^2\) is equal to:
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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..........
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The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40 . Then the correct variance is:
[JEE Main 2023, 15 Apr (Shift 1)]
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The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35 . If \(\alpha\) and \(\sqrt{\beta}\) are the mean and standard deviation respectively for correct data, then \((\alpha, \beta)\) is:
[JEE Main 2021, 26 Aug (Shift 1)]
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The mean of 10 numbers
\(7 \times \ 8\ ,10 \times \ 10\ \ ,13 \times \ 12\ \ ,16 \times \ 14\ \ ,\ ....\ is\)
[JEE Main 2021, 31 Aug (Shift 1)]
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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____________
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Let the mean of 6 observation \(1,2,4,5, x\) and \(y\) be 5 and their variance be 10 . Then their mean deviation about the mean is equal to
[JEE Main 2023, 11 Apr (Shift 2)]
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For the frequency distribution:
Variate \((x): x_1 x_2 x_3 \ldots x_{15}\)
Frequency \((f): f_1 f_2 f_3 \ldots f_{15}\) Where \(0
[JEE Main 2020, 3 Sep (Shift 1)]
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If the mean and variance of six observations \(7,10,11,15\), \(a, b\) are 10 and \(\frac{20}{3}\), respectively, then the value of \(|a-b|\) is equal to
[JEE Main 2021, 20 Jul (Shift 2)]
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If the mean and the standard deviation of the data \(3,5,7\), \(a, b\) are 5 and 2 respectively, then \(a\) and \(b\) are the roots of the equation:
[JEE Main 2020, 5 Sep (Shift 2)]
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Let \(n\) be an odd natural number such that the variance of \(1,2,3,4, \ldots n\) is 14 . Then \(n\) is equal to___________
[JEE Main 2021, 27 Aug (Shift 1)]
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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
[JEE Main 2023, 6 Apr (Shift 1)]
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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 \(m\) and \(n\) are co-prime, then \(m+n\) is equal to
[JEE Main 2023, 08 Apr (Shift 2)]
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Let \(x_i(1 \leq i \leq 10)\) be ten observations of a random variable \(X\). If \(\sum_{i=1}^{10}\left(x_i-p\right)=3\) and \(\sum_{i=1}^{10}\left(x_i-p\right)^2=9\) where \(p \neq 0\) \(p \in R\), then the standard deviation of these observations is
[JEE Main 2020, 3 Sep (Shift 2)]
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The mean and standard deviation of \(10\) observations are \(20\) and \(8\) respectively. Later on, it was observed that one observation was recorded as \(50\) instead of \(40\) .Then the correct variance is:
[JEE Main 2023, 15 Apr (Shift 1)]
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Let the observations \(x_i(1 \leq i \leq 10)\) satisfy the equations, \(\sum_{i=1}^{10}\left(x_i-5\right)=10\) and \(\sum_{i=1}^{10}\left(x_i-5\right)^2=40\). If \(\mu\) and \(\lambda\) are the mean and the variance of the observations, \(x_1-3, x_2-3\), \(\ldots x_{10}-3\), then the ordered pair \((\mu, \lambda)\) is equal to:
[JEE Main 2020, 9 Jan (Shift 1)]
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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:
[JEE Main 2023, 25 Jan (Shift 1)]
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Let \(9=x_1 [JEE Main 2023, 1 Feb (Shift 2)]
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The mean and variance of \(7\) observations are \(8\) and \(16\) respectively. If two observations are \(6\) and \(8\), then the variance of the remaining \(5\) observations is:
[JEE Main 2021, 31 Aug (Shift 2)]
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Let the six numbers \(a_1, a_2, a_3, a_4, a_5, a_6\) be in A.P. and \(a_1+a_3=10\). If the mean of these six numbers is \(\frac{19}{2}\) and their variance is \(\sigma^2\), then \(8 \sigma^2\) is equal to
[JEE Main 2023, 24 Jan (Shift 2)]
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Let the mean and standard deviation of marks of class \(A\) of 100 students be respectively 40 and \(\alpha(>0)\), and the mean and standard deviation of marks of class \(B\) of \(n\) students be respectively 55 and \(30-\alpha\). If the mean and variance of the marks of the combined class of \(100+n\) students are respectively 50 and 350 , then the sum of variances of classes \(A\) and \(B\) is:
[JEE Main 2023, 31 Jan (Shift 2)]
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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:
[JEE Main 2023, 25 Jan (Shift 1)]
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Let \(S\) be the set of all values of \(a_1\) for which the mean deviation about the mean of 100 consecutive positive integers \(a_1, a_2, a_3, \ldots . ., a_{100}\) is 25 . Then \(S\) is
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The mean of 6 distinct observations is 6.5 and their variance is 10.25 . If 4 out of 6 observations are \(2,4,5\) and 7 , then the remaining two observations are :
[JEE Main 2021, 20 Jul (Shift 1)]
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