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Exercise 11.6 · Q12

Q.If XX is a binomial random variable with expected value 66 and variance 2.42.4, then P(X=5)P(X=5) is

(1) (105)(35)6(25)4\dbinom{10}5\left(\dfrac35\right)^6\left(\dfrac25\right)^4
(2) (105)(35)10\dbinom{10}5\left(\dfrac35\right)^{10}
(3) (105)(35)4(25)6\dbinom{10}5\left(\dfrac35\right)^4\left(\dfrac25\right)^6
(4) (105)(35)5(25)5\dbinom{10}5\left(\dfrac35\right)^5\left(\dfrac25\right)^5.
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Recover n,pn,p from the given mean and variance exactly as in the mean/variance-of-standard-distributions technique, then substitute k=5k=5 into the binomial pmf.

Step 1. Set up the two equations. np=6np=6 and npq=2.4npq=2.4.

Step 2. Solve for qq, then pp, then nn. q=npqnp=2.46=0.4⇒p=0.6q=\dfrac{npq}{np}=\dfrac{2.4}6=0.4\Rightarrow p=0.6. Then n=60.6=10n=\dfrac6{0.6}=10.

Step 3. Convert p,qp,q to fractions. p=0.6=35, q=0.4=25p=0.6=\dfrac35,\ q=0.4=\dfrac25. …

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