Question

In a binomial distribution, n = 12 and π = .60.

Answer 1

For binomial distribute we use formula

$nCr (p)^r (q)^{n-r}$

the probability mass function of the binomial distribution is:

$nCr (\pi)^r (1-\pi)^{n-r}$

Given n = 12 and π = .6. (Replace the values)

the probability for x = 5, we need to find P(x=5). Here r= 5

P(x=5)= $12C5(\pi )^5 (1-\pi)^{12-5}$

= $12C5(0.6 )^5 (0.4)^{7}$= 0.10090 = 0.101

(b) the probability for x ≤ 5

P(x<=5) = P(x=0) + P(x=1) +P(x=2)+ P(x=3) + P(x=4) + P(x=5)

P(x<=5) = $12C0(0.6 )^0(0.4)^{12}$ + $12C1(0.6 )^1(0.4)^{11}$ +  $12C2(0.6 )^2(0.4)^{10}$ + $12C3(0.6 )^3(0.4)^{9}$ + $12C4(0.6 )^4(0.4)^{8}$ + $12C5(0.6 )^5 (0.4)^{7}$

= 0.15821229 = 0.158

(c) find the probability for x ≥ 6

P(x>=6) =P(x=6) + P(x=7) +P(x=8)+ P(x=9) + P(x=10) + P(x=11)+ P(x=12)

P(x>=6) = $12C6(0.6 )^6(0.4)^{6}$ + $12C7(0.6 )^7(0.4)^{5}$ + $12C8(0.6 )^8(0.4)^{4}$ + $12C9(0.6 )^9(0.4)^{3}$ + $12C10(0.6 )^{10}(0.4)^{2}$ +   $12C11(0.6 )^{11}(0.4)^{1}$  +  $12C12(0.6 )^{12}(0.4)^{0}$

= 0.8417877 = 0.0842 (rounded to 3 decimal places)