The outcome of each of 30 items was observed; 10 items gave an outcome {1 \over 2} – d each, 10 items gave outcome {1 \over 2} each and the remaining 10 items gave outcome {1 \over 2}+ d each. If the variance of this outcome data is {4 \over 3} then |d| equals :

If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -

A data consists of n observations : x1, x2, . . . . . . ., xn. If $$\sum _{i=1}^n{\left(x_i+1\right)}^2=9n$$ and $$\sum _{i=1}^n{\left(x_i-1\right)}^2=5n,$$ then the standard deviation of this data is :

5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :

The sum of all natural numbers 'n' such that 100 1 is :

The sum \sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} is equal to

If three distinct numbers a, b, c are in G.P. and the equations ax2 + 2bx + c = 0 and dx2 + 2ex + ƒ = 0 have a common root, then which one of the following statements is correct?

Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ..... be 50n + {{n(n - 7)} \over 2}A, where A is a constant. If d is the common difference of this A.P., then the ordered pair (d, a50) is equal to

Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are addded to the total number of balls used in forming the equilaterial triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is :-

If the sum and product of the first three term in an A.P. are 33 and 1155, respectively, then a value of its 11th term is :-

The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +.... upto 11th term is :-

The sum {{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + ..... upto 10 terms is:

If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to :

Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$\ne$$ 0 and 0 < r $$\le$$ {1 \over 2} . If 3$$a$$, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :

Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the product a1a4a5, is :

For x $$\epsilon$$ R, let [x] denote the greatest integer $$\le$$ x, then the sum of the series \left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right] is :

Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6= – 48, then S10 is equal to :

If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :

The sum 1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}- {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right) is equal to :