AP ICET 2013 Question Paper

when we calculated mean we ascending data 

here if we take -24 instead of the 24 so it comes in first place as it is smallest value and affect first 3 value of 6 observation 

so (24-(-24))\3 = 16 new mean      Answer 

Since 8 is less than 32 and 32 is more than 6, so the middle value remains unchanged and hence the median of new data  observations taken together will be 32. 

  Answer 

Mode is the observation that occurs most frequently in the data 

here 13 is repeated 3 times so 13 is our answer 

The standard deviation of n natural numbers = t 

Therefore, but we know that 

First n natural numbers are 1,2,3,…,n.

Mean=(1+2+3+…+n)÷n

=[n(n+1)/2]/n

(n+1)/2 …(i)

And, 1²+2²+3²+…+n²=n(n+1)(2n+1)/6

Or, (1²+2²+3²+…+n²)/n= {(n+1)(2n+1)/6} …(ii)

Variance of first n natural numbers $$\sqrt{t}$$  = {(1²+2²+3²+…+n²)/n} - (Mean)²

= {(n+1)(2n+1)/6}-{(n+1)²/4}

= (n+1)/12{4n+2-3(n+1)}

= (n+1)(n-1)/12

$$\sqrt{t}$$ = (n²-1)/12

and we have the equation given 

$$12t^{2} + 1 =$$ 

t^2 = -1\12  put the of t we get 

n^2 Answer 

Let say

Y = 2X + 17

Taking variance of both sides:

V(Y) = V(2X) + V(17)
V(Y) = 2 V(X) + 0
V(Y) = 2 * 12
V(Y) = 24                              Answer 

V(aX) = a^2 * V(X) .. where a is a constant

V(a) = 0 .. where a is a constant

when the 3 cion are tossed the total number of cases are = $$2^3$$ = 8 

and at least two heads are there in the following = HHT, HTH, THH, HHH = 4 

then the probility of getting at least two heads  = number of favourable cases \ total number of cases 

then the probility of getting at least two heads = 4\8 = 1\2 Answer 

when the two unbiased dice are thrown the number of total  cases are = 36 

let the both the dice show the same number then cases are = (1,1), (2,2), (3,3),(4,4),(5,5) and (6,6) = 6 

then the probility of getting different number = 1 - probility of getting same number on both dice 

then the probility of getting different number= 1 - 6\36

then the probility of getting different number = 1-1\6 

then the probility of getting different number = 5\6 Answer 

 If X1, X2, ... , Xn are mutually independent normal random variables with means μ1, μ2, ... , μn and variances σ21,σ22,⋯,σ2nσ12,σ22,⋯,σn2, then the linear combination 

of ranks  $$X_i$$ and $$Y_i$$ means have equal value 4 and 4 as i = 1,2,3,4......8 

    Y =  $$\sum_{i = 1}^{8} d_i^{2} $$ = 21 

Then, finding the probability that X is greater than Y reduces to a normal probability calculation:

P(X>Y )= P(X − Y > 0) = P( Z > 0 − 1\4 ) = P(Z>−0.75 ) = P(Z<0.75) = 0.75 Answer 

Given that 5 boys and 4  girls sit in a row at random.

Total number of ways in which 9 persons can be arranged in a row = 9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1

=  5 boys can be arranged themselves in  5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.

 B 1 B 2 B 3 B 4 B 

Now,

Then there are 4 positions left(1,2,3,4,).

The 4 girls can be seated in (1,2,3,4) =  4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1  ways.

The girls can be arranged themselves in   4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.

Therefore, the number of ways both girls and boys can sit alternatively = 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 $$\times$$   4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1

Hence,

Required probability = 

(5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 $$\\times 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)\( 9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)

= (120 $$\times$$ 24)\(9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)

=  4/504

=  1/126.    Answer 

The question is asking the probability of getting "exactly" 52 Sundays and not at least 52 Sundays.

There are 52 weeks in a year, so there will be 52 Sundays, no doubts  about it. However, 52 $$\times$$ 7 = 364 days  

 so that still leaves a day for a normal year, and two days for a leap year.

Now, let us look at the combinations of the two days possible.

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

Now, to get exactly 52 Sundays, none of the last two days should be a Sunday. As you can see from the combinations listed above, that is possible in 5 out of 7 cases. So the Probability would be 5/7.       Answer