The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
2
2.57
3
3.75
Mean deviation for \( n \) observations \( x_1, x_2, ..., x_n \) from their mean \( \bar{x} \) is given by
\( \sum (x_i - \bar{x}) \)
\( \dfrac{1}{n} \sum |x_i - \bar{x}| \)
\( \sum (x_i - \bar{x})^2 \)
\( \dfrac{1}{n} \sum (x_i - \bar{x})^2 \)
The lives (in hours) of 5 bulbs were noted as: 1357, 1090, 1666, 1494, 1623. The mean deviation (in hours) from their mean is
178
179
220
356
Following are the marks obtained by 9 students in a mathematics test: 50, 69, 20, 33, 53, 39, 40, 65, 59. The mean deviation from the median is
9
10.5
12.67
14.76
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is
\( \sqrt{\dfrac{52}{7}} \)
\( \dfrac{52}{7} \)
\( \sqrt{6} \)
6
Let \( x_1, x_2, ..., x_n \) be \( n \) observations and \( \bar{x} \) be their arithmetic mean. The formula for the standard deviation is
\( \sum (x_i - \bar{x})^2 \)
\( \dfrac{\sum (x_i - \bar{x})^2}{n} \)
\( \sqrt{\dfrac{\sum (x_i - \bar{x})^2}{n}} \)
\( \sqrt{\dfrac{\sum x_i^2}{n} + \bar{x}^2} \)
The mean of 100 observations is 50 and the standard deviation is 5. The sum of all squares of all the observations is
50000
250000
252500
255000
Let \( a, b, c, d, e \) be observations with mean \( m \) and standard deviation \( s \). The standard deviation of the observations \( a+k, b+k, c+k, d+k, e+k \) is
s
ks
s+k
\( \dfrac{s}{k} \)
Let \( x_1, x_2, x_3, x_4, x_5 \) be 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
k + s
\( \dfrac{s}{k} \)
ks
s
Let \( x_1, x_2, ..., x_n \) be observations. Let \( w_i = lx_i + k \). If the mean of \( x_i \) is 48 and SD = 12, and the mean of \( w_i \) is 55 and SD = 15, values of \( l \) and \( k \) are
\( l = 1.25, k = -5 \)
\( l = -1.25, k = 5 \)
\( l = 2.5, k = -5 \)
\( l = 2.5, k = 5 \)
Standard deviation for the first 10 natural numbers is
5.5
3.87
2.97
2.87
Consider the numbers 1 to 10. If 1 is added to each number, the variance of the numbers so obtained is
6.5
2.87
3.87
8.25
Consider the first 10 positive integers. If each number is multiplied by \(-1\) and then 1 is added, the variance becomes
8.25
6.5
3.87
2.87
A sample of size 60 has \( \sum x = 960 \) and \( \sum x^2 = 18000 \). The variance is
6.63
16
22
44
Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviations is
0
1
1.5
2.5
The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be
81
57
36
25