The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

Let xi (1 $$\le$$ i $$\le$$ 10) be ten observations of a random variable X. If $$\sum _{i=1}^{10}\left(x_i-p\right)=3$$ and $$\sum _{i=1}^{10}{\left(x_i-p\right)}^2=9$$ where 0 $$\ne$$ p $$\in$$ R, then the standard deviation of these observations is :

Let the observations xi (1 $$\le$$ i $$\le$$ 10) satisfy the equations, $$\sum _{i=1}^{10}\left(x_1-5\right)$$ = 10 and $$\sum _{i=1}^{10}{\left(x_1-5\right)}^2$$ = 40. If $$\mu$$ and $$\lambda$$ are the mean and the variance of the observations, x1 – 3, x2 – 3, ...., x10 – 3, then the ordered pair ($$\mu$$, $$\lambda$$) is equal to :

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is

The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 resepectively. Each of these 10 observations is multiplied by p and then reduced by q, where p $$\ne$$ 0 and q $$\ne$$ 0. If the new mean and new s.d. become half of their original values, then q is equal to

For the frequency distribution : Variate (x) : x1 x2 x3 .... x15 Frequency (f) : f1 f2 f3 ...... f15 where 0 0, the standard deviation cannot be :

If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :

If the sum of the series 20 + 19{3 \over 5} + 19{1 \over 5} + 18{4 \over 5} + ... upto nth term is 488 and the nth term is negative, then :

If 1+(1–22.1)+(1–42.3)+(1-62.5)+......+(1-202.19)= $$\alpha$$ - 220$$\beta$$, then an ordered pair $$\left(\alpha ,\beta \right)$$ is equal to:

The minimum value of 2sinx + 2cosx is :

Let a1, a2, ..., an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $$\le$$ n $$\le$$ 50, then the ordered pair (Sn-4, an–4) is equal to:

If 210 + 29.31 + 28 .32 +.....+ 2.39 + 310 = S - 211, then S is equal to :

If $$3^{2\sin 2\alpha -1}$$, 14 and $$3^{4-2\sin 2\alpha }$$ are the first three terms of an A.P. for some $$\alpha$$, then the sixth terms of this A.P. is:

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

If the sum of the first 20 terms of the series $${\log }_{\left(7^{1/2}\right)}x+{\log }_{\left(7^{1/3}\right)}x+{\log }_{\left(7^{1/4}\right)}x+...$$ is 460, then x is equal to :

Let a , b, c , d and p be any non zero distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :

The common difference of the A.P. b1, b2, … , bm is 2 more than the common difference of A.P. a1, a2, …, an. If a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :

Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -{1 \over 2} , then the greatest number amongst them is:

Let S be the sum of the first 9 terms of the series : {x + k$$a$$} + {x2 + (k + 2)$$a$$} + {x3 + (k + 4)$$a$$} + {x4 + (k + 6)$$a$$} + .... where a $$\ne$$ 0 and x $$\ne$$ 1. If S = {{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}, then k is equal to :

Let $$a_1$$ , $$a_2$$ , $$a_3$$ ,....... be a G.P. such that $$a_1$$ < 0, $$a_1$$ + $$a_2$$ = 4 and $$a_3$$ + $$a_4$$ = 16. If $$\sum _{i=1}^9{a}_i=4\lambda$$, then $$\lambda$$ is equal to: