1
JEE Main 2022 (Online) 26th June Morning Shift
+4
-1

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 :

A
60
B
55
C
50
D
45
2
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
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 :
A
$${{92} \over 5}$$
B
$${{134} \over 5}$$
C
$${{536} \over {25}}$$
D
$${{112} \over 5}$$
3
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
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 :
A
(11, 26)
B
(10.5, 25)
C
(11, 25)
D
(10.5, 26)
4
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
Let the mean and variance of the frequency distribution

$$\matrix{ {x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \cr {f:} & 4 & 4 & \alpha & \beta \cr }$$

be 6 and 6.8 respectively. If x3 is changed from 8 to 7, then the mean for the new data will be :
A
4
B
5
C
$${{17} \over 3}$$
D
$${{16} \over 3}$$
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