PYQSidePowered by
ExamsMy practice
Let the mean and the variance of 6 observations a, b, 68, 44, 48, 60 be 55 and 194, respectively. If a>b, then a+3b is
(1) 180
Mean: a+b+2206=55 a+b=110
Variance: (xi-55)26=194 (a-55)2+(b-55)2+169+121+49+25=1164
(a-55)2+(b-55)2=800
∵ a+b=110, put a-55=t, b-55=-t 2t2=800 t=20 ∵ a>b
a=75, b=35
a+3b=75+105=180
Answer: (1)
Let M denote the median of the following frequency distribution [figure] Then 20M is equal to :
If the mean and variance of the data 65,68,58,44,48,45,60, , ,60 where > , are 56 and 66.2 respectively, then 2+ 2 is equal to .
If the mean and variance of five observations are 245 and 19425 respectively and the mean of the first four observations is 72, then the variance of the first four observations is equal to
Let a1, a2, a10 be 10 observations such that k=110 ak=50 and k<j ak aj=1100. Then the standard deviation of a1, a2, , a10 is equal to :
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 and 2 denote the mean and variance of the correct observations respectively, then 15 ( + 2+ 2 ) is equal to .