Question 464537: Q1-A certain binomial distibution has mean 4 and varince 3.Find for this distribution, the probability of obtaining
(i)exactly two successes (ii)no successes (iii)more than three successes and (iv)lesser than 3 successes>
Q2-Assuming the mean height of soliders to be 68.22 inches with a varince of 10.8, (i) find how many soliders in a regiment of 1000 would you expect to be cover 6 feet tall.
(ii) find the probability that a solider chosen at random is shorter than 60 inches.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A certain binomial distibution has mean 4 and variance 3.
np = 4
npq = 3
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Dividing (npq)/(np) you get q = 3/4
Then p = 1/4
And n must be 16
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Find for this distribution, the probability of obtaining:
(i)exactly two successes
P(x = 2) = 16C2(1/4)^2(3/4)^14 = 0.1336
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(ii)no successes
P(x = 0) = (3/4)^16 = 0.0100
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(iii)more than three successes
P(4<= x <=16) = 1 - P(0<= x <=3) = 1 - binomcdf(16,1/4,3) = 0.5950
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(iv)lesser than 3 successes
P(0<= x <=2) = 0.1971
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Q2-Assuming the mean height of soliders
to be 68.22 inches with a variance of 10.8,
(i) Find how many soliders in a regiment of 1000
would you expect to be over 6 feet tall.
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z(72) = (72-68.22)/sqrt(10.8) = 1.2841
P(x > 72) = P(z > 1.2841) = 0.0996
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Expected number of soldiers over 6 ft. = 0.0996* 1000 = 100 when rounded up.
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(ii) find the probability that a solider chosen at random is shorter than 60 inches.
Use the same process as used above.
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Cheers,
Stan H.
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