Suppose a population A has 100 observations 101, 102, …, 200, and another population B has 100 observations 151, 152, …, 250. If \(V_A\) and \(V_B\) represent the variances of the two populations, respectively, then \(\dfrac{V_A}{V_B}\) is
Let \(\bar{x}\), \(M\) and \(\sigma^2\) be respectively, the mean, mode and variance of \(n\) observations \(x_1, x_2, ..., x_n\) and \(d_i = -x_i - a\), \(i = 1, 2, ..., n\), where \(a\) is any number.Statement-1: Variance of \(d_1, d_2, ..., d_n\) is \(\sigma^2\).Statement-2: Mean and mode of \(d_1, d_2, ..., d_n\) are \(-\bar{x} - a\) and \(-M - a\), respectively.
Runs scored by a batsman in 10 innings are: 38, 70, 48, 34, 42, 55, 63, 46, 54, 44. The mean deviation is
The sum of 100 observations and the sum of their squares are 400 and 2474, respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is
Let $x_1, x_2, x_3, \ldots, x_k$ be $k$ observations and $w_i = ax_i + b$ for $i = 1, 2, 3, \ldots, k$, where $a$ and $b$ are constants. If mean of $x_i$ is 52 and their standard deviation is 12 and mean of $w_i$ is 60 and their standard deviation is 15, then the value of $a$ and $b$ should be
Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. If 1 is added to each number, the variance of the numbers so obtained is
$V_1$ = variance of {13, 16, 19, . . . , 103}. $V_2$ = variance of {3, 6, 9, . . . , 93}. Find $\frac{V_1}{V_2}$.
Let \(x_1, x_2, \ldots, x_n\) be \(n\) observations. Let \(w_i = lx_i + k\) for \(i = 1, 2, \ldots, n\), where \(l\) and \(k\) are constants. The mean of \(x_i\)'s is 48 and their standard deviation is 12. Also, the mean of \(w_i\)'s is 55 and standard deviation of \(w_i\)'s is 15. The values of \(l\) and \(k\) should be
For the data $x: 1,3,5,7,9$; frequency $4,24,28,\alpha,8$ with mean 5, then $\dfrac{3\alpha}{m+\sigma^2}$ is equal to _______.
Consider the following statements regarding the data set \(2x_1, 2x_2, \ldots, 2x_n\) where \(\sigma^2\) is the variance of \(x_1, x_2, \ldots, x_n\):Statement-1: The variance of \(2x_1, 2x_2, \ldots, 2x_n\) is \(4\sigma^2\).Statement-2: The AM of \(2x_1, 2x_2, \ldots, 2x_n\) is \(2\bar{x}\).Which of the following is correct?
Let \(x_1, x_2, \ldots, x_n\) be \(n\) observations, and let \(\bar{x}\) be their arithmetic mean and \(\sigma^2\) be the variance.Statement 1: Variance of \(2x_1, 2x_2, \ldots, 2x_n\) is \(4\sigma^2\).Statement 2: Arithmetic mean of \(2x_1, 2x_2, \ldots, 2x_n\) is \(4\bar{x}\).
For the distribution $X_i: 0,1,2,3,4,5$; $f_i: k+2,2k,k^2-1,k^2-1,k^2+1,k-3$ with $\sum f_i=62$, then $[\mu^2+\sigma^2]$ is equal to
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 ______.