Statistics Questions (255)

The following information relates to a sample of size 60: \(\Sigma x^2 = 18000\), \(\Sigma x = 960\). The variance is
If a variate X is expressed as a linear function of two variates U and V in the form \(X = aU + bV\), then the mean \(\bar{X}\) of X is
Let $X = \{11, 12, 13, \ldots, 40, 41\}$ and $Y = \{61, 62, 63, \ldots, 90, 91\}$ 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 $|\bar{x} + \bar{y} - \sigma^2|$ is equal to ______.
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is
The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the new variance would be
Five positive numbers in GP have mean $\frac{31}{10}$ and the sum of reciprocals is $\frac{31}{40}$. If variance $=\frac{m}{n}$ (coprime), then $m+n$ is equal to
The weighted mean of the first \(n\) natural numbers whose weights are equal to the squares of the corresponding numbers is
Let \(x_1, x_2, x_3, x_4, x_5\) be the observations with mean \(m\) and standard deviation \(s\). The standard deviation of the observations \(kx_1, kx_2, kx_3, kx_4, kx_5\) is
In an experiment with 15 observations on \(x\), then following results were available: \(\sum x^2 = 2830\), \(\sum x = 170\). One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the correct variance is
For the frequency distribution (class: 0-10, 10-20, 20-30, 30-40, 40-50; frequency: 2,3,x,5,4) with mean 28, the variance is ________.
The mean and the SD of 1, 2, 3, 4, 5, 6 are
If for some \(x \in R\), the frequency distribution of the marks obtained by 20 students in a test isMarks2357Frequency\((x+1)^2\)\(2x-5\)\(x^2-3x\)\(x\)then the mean of the marks is ______.
The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then \(\frac{y}{x}\) is equal to
If the mean of the set of numbers \(x_1, x_2, x_3, \ldots, x_n\) is \(\bar{x}\), then the mean of the numbers \(x_i + 2i\), \(1 \le i \le n\) is
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 ______.
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:
Total number of terms = 34. So mean of 17th and 18th term is median. Find the median if $x_{17} = 130, x_{34} = 134$.
Mean and SD of 10 observations are 20 and 8. One observation was recorded as 50 instead of 40. Then the correct variance is
For a normal distribution if the mean is \(M\), mode is \(M_0\) and median is \(M_d\), then
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is:
Consider the frequency distribution, where A is a positive integer:\(x\)\(A\)\(2A\)\(3A\)\(4A\)\(5A\)\(6A\)\(f\)211111If the variance is 160, then the value of A is
The mean of the numbers \(a, b, 8, 5\), and 10 is 6 and the variance is 6.80. Then which one of the following gives the possible values of \(a\) and \(b\)?
If the mean deviation about the median of the numbers \(a, 2a, \ldots, 50a\) is 50, then \(|a|\) equals
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 $15(\mu+\mu^2+\sigma^2)$ is equal to
Mean and variance of 8 numbers $x,y,10,12,6,12,4,8$ are 9 and 9.25. If $x>y$, then $3x-2y$ is equal to _______
If the variance of the first $n$ natural numbers is 10 and the variance of the first $m$ even natural numbers is 16, then the value of $m + n$ is equal to
In a moderately skewed distribution, the values of mean and median are 5 and 6, respectively. The value of mode in such a situation is approximately equal to
If both the mean and the standard deviation of 50 observations \(x_1, x_2, \ldots, x_{50}\) are equal to 16, then the mean of \((x_1 - 4)^2, (x_2 - 4)^2, \ldots, (x_{50} - 4)^2\) is ______.
For observations $x_1, x_2, \ldots, x_{30}$, if Mean $= 16$ and $\sum x_i^2 = 8000$, find the variance and then find the mean value of $(x_1 - 4)^2 + (x_2 - 4)^2 + \cdots + (x_{30} - 4)^2$.
The mean and variance of a data set comprising 15 observations are 15 and 5 respectively. If one of the observation 15 is deleted and two new observations 9 and 8 are added to the data, then the new variance of resulting data is
If the mean and the variance of the data (Class 4–8, 8–12, 12–16, 16–20; Frequency 3, $\lambda$, 4, 7) are $\mu$ and 19 respectively, then the value of $\lambda+\mu$ is
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$, then $96\sigma^2$ is equal to _____
Let $\alpha,\beta\in\mathbb{R}$. Let the mean and the variance of 6 observations $-3,4,7,-6,\alpha,\beta$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:
For 50 observations \(a, 2a, 3a, \ldots, 50a\), the mean deviation about the median is minimized and \(\dfrac{1}{n}\sum|x_i - A|\) is minimized when \(A\) is the median. If \(625a = 2500\), find \(a\).
The average of \(n\) numbers \(x_1, x_2, x_3, \ldots, x_n\) is \(M\). If \(x_n\) is replaced by \(x'\), then new average is
Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\displaystyle\sum_{i=1}^{10}(x_i - 2) = 30$, $\displaystyle\sum_{i=1}^{10}(x_i - \beta)^2 = 98$, $\beta > 2$, and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, \ldots, 2(x_{10} - 1) + 4\beta$, then $\frac{\beta\mu}{\sigma^2}$ is equal to:
For a statistical data $x_{1},x_{2},\dots,x_{10}$ 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:
The mean of \(n\) items is \(\bar{X}\). If the first item is increased by 1, second by 2 and so on, then the new mean is
The following data give the distribution of heights of students:Height (in cm)160150152161156154155Number of students12844337The median of the distribution is
The mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50, respectively. The correct mean is
If the arithmetic mean of the numbers \(x_1, x_2, x_3, \ldots, x_n\) is \(\bar{x}\), then the arithmetic mean of the numbers \(ax_1 + b,\ ax_2 + b,\ ax_3 + b,\ \ldots,\ ax_n + b\), where \(a, b\) are two constants, would be
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
If the mean and variance of 6 observations are 8 and 16 respectively, and 4 of the observations are $6,7,14,8$, find the remaining two observations.
In an experiment with 15 observations on \(x\), the following results were available: \(\Sigma x^2 = 2830\), \(\Sigma x = 170\). One observation, 20, was found to be wrong and was replaced by the correct value 30. Then the corrected variance is
Consider data on $X$ taking values $0, 2, 4, 8, \ldots, 2^n$ with frequencies ${}^nC_0, {}^nC_1, \ldots, {}^nC_n$ respectively. If the mean of this data is $\dfrac{728}{2^n}$, then $n$ is equal to
Let \(r\) be the range and \(S^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2\) be the SD of a set of observations \(x_1, x_2, \ldots, x_n\), then
The variance of the data 2, 4, 6, 8, 10 is
Consider the first 10 positive integers. If we multiply each number by \(-1\) and then add 1 to each number, the variance of the numbers so obtained is
Frequency distribution with classes $[0,10),[10,20),\ldots,[50,60)$ and frequencies $3,c,d,11,5,5$ has mean $31$ and median $340/11$. Value of $\left[\tan^{-1}\!\dfrac{2cd}{d^2-c^2}\right]$ (GIF) is
The mean weight of 9 items is 15. If one more item is added to the series, the mean becomes 16. The value of the $10^{\text{th}}$ item is