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Exercise 11.5 · Q7

Q.The mean and standard deviation of a binomial variate XX are respectively 66 and 22. Find

(i) the probability mass function
(ii) P(X=3)P(X=3)
(iii) P(X≥2)P(X\ge2).
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From mean =np=6=np=6 and variance =(sd)2=npq=4=(\text{sd})^2=npq=4, dividing gives qq directly, then pp and nn; the pmf, a specific value, and a tail probability follow.

Step 1. Set up the two equations. np=6np=6 and npq=(sd)2=22=4npq=(\text{sd})^2=2^2=4.

Step 2. Solve for qq, then pp, then nn. Dividing: q=npqnp=46=23q=\dfrac{npq}{np}=\dfrac46=\dfrac23, so p=1−23=13p=1-\dfrac23=\dfrac13. Then n=npp=61/3=18n=\dfrac{np}{p}=\dfrac6{1/3}=18.

Step 3. (i) Write the pmf. X∼B ⁣(18,13)X\sim B\!\left(18,\dfrac13\right): f(x)=(18x)(13)x(23)18−xf(x)=\dbinom{18}x\left(\dfrac13\right)^x\left(\dfrac23\right)^{18-x}, x=0,1,…,18x=0,1,\dots,18.

Step 4. (ii) P(X=3)P(X=3). P(X=3)=(183)(13)3(23)15=816×127×3276814348907=816×32768387420489=26738688387420489≈0.0690P(X=3)=\dbinom{18}3\left(\dfrac13\right)^3\left(\dfrac23\right)^{15}=816\times\dfrac1{27}\times\dfrac{32768}{14348907}=\dfrac{816\times32768}{387420489}=\dfrac{26738688}{387420489}\approx0.0690.

Step 5. (iii) P(X≥2)=1−P(X=0)−P(X=1)P(X\ge2)=1-P(X=0)-P(X=1). …

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