Please help<q.a.>

CPT 829 views 5 replies

Q1In a election the number of candidates is one more than the no of members to be elected.If a voter can vote in254 ways, find the number of candidates.

a,8 b,10 c,7 d, none

Q2From 17 consonants and 5 vowels,how many words of 3 consonants and two vowels can be made if all the letters are different.

a 810000 b 816000 c 815000 d none

Q3A boat is to be manned by 8 men of which 3 can row  only one side and two only on the other.In how many ways can the crew be arranged?

a 1720 b 7700 c 1728 d none

Q4For a normal distribution,the first moment about origin is 35 and the second moment about is 10.Find the four central moments

a 5,10,0,100 b 10,6,0,200 c 0,10,0,300 d,none

Q5 Find the sample size such that the probability of the sample means differing from the population mean nby not more than 1/10th of the S.D. IS 0.95.

a 300 b 384 c 395 d none

Replies (5)

Q2 )  From 17 consonents 3 consonents are to be taken and from 5 vowels 2 vowels are to be taken. The resulting letters from a word. Hence the order is also important. Hence we use Permutation.

   17P3 * 5P2  = (17!/14!) * (5!/3!) = 15*16*17*4*5 = 81600  -- Ans..

q.3. 8c3*8c2 = 56*28 = 1568.

answer none.

1. let the number of of candiates be n. now number of ways voter can vote is ncn-1 + ncn-2 +...+nc1 = 2^n - ncn-nc0 = 2^n - 2so 2^n = 256 imlies n = 8

3. in the 3 remainging men the one person to be roweing on the side with 3 persons can be selected in 3 ways now we got 4 persons on each side we can arrange them in 4! ways so answer is 3*4!*4! = 1728

q.1 i can't understand.

4.the first and third cental moments of a normal distribution are always 0. the second moment is variance and 4th moment is 3*variance^2  variance = E(X^2) - (E(X))^2 in the given quesiton variance is coming to be negative check the figures once again.

1. the voter can vote for all the positions or one less than all the positions or two less than all the positions or ........ or only one positon to select poeple for n-1 positions ncn-1, n-2 positions ncn-2 so on then apply the fact that sum of all binomial coefficients is 2^n;

 


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