1
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be $${24 \over 5}$$ and $${194 \over 25}$$ respectively. If the mean and variance of the first 4 observation are $${7 \over 2}$$ and a respectively, then (4a + x5) is equal to:

A
13
B
15
C
17
D
18
2
JEE Main 2022 (Online) 27th June Evening Shift
+4
-1

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $${9 \over 4}$$ respectively. Then $${x^4} + {y^2}$$ is equal to :

A
162
B
320
C
674
D
420
3
JEE Main 2022 (Online) 26th June Evening Shift
+4
-1

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :

A
10
B
36
C
43
D
60
4
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
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