Let X represent a binomial random variable with n = 170 and p = 0.6. Find the following probabilities. (Do not round intermediate calculations. Round your final answers to 4 decimal places.)
Probability
a.P(X ≤ 100)
b.P(X > 110)
c.P(105 ≤ X ≤ 115)
d.P(X = 90)
(A) To find
Using binomcdf(n,p,k)
it is given that n = 170, p=0.6 and k = 100
= binomcdf(170,0.6,100)
= 0.4053
Therefore, = 0.4053
(B) To find
Using P(x more than 110)= 1 - binomcdf(n,p,k)
it is given that n = 170, p=0.6 and k = 110
= 1 - binomcdf(170,0.6,110)
= 1 - 0.9091
= 0.0909
Therefore, = 0.0909
(C) To find
We can write it as
this implies
= binomcdf (170,0.6,115) - binomcdf ( 170, 0.6, 105-1)
= binomcdf (170,0.6,115) - binomcdf ( 170, 0.6, 104)
= 0.9837 - 0.6505
= 0.3332
Therefore, = 0.3332
(D) To find P(x = 90)
Using binompdf(n,p,k)
it is given that n = 170, p=0.6 and k = 90
= binompdf(170,0.6,90)
= 0.0108
Therefore,P(x=90) = 0.0108
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