Question

Consider a binomial probability distribution with pequals=0.3 and nequals=8. What is the probability of the following?...

Consider a binomial probability distribution with

pequals=0.3

and

nequals=8.

What is the probability of the following?

a)

exactly three successes

b)

less than three successes

c)

sixsix

or more successes

a)

Upper P left parenthesis x equals 3 right parenthesisP(x=3)equals=nothing

(Round to four decimal places as needed.) b)

Upper P left parenthesis x less than 3 right parenthesisP(x<3)equals=nothing

(Round to four decimal places as needed.) c)

Upper P left parenthesis x greater than or equals 6 right parenthesisP(x≥6)equals=nothing

(Round to four decimal places as needed.)

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Answer #1

Binomial distribution: P(X) = nCx px qn-x

n = 8

p = 0.3

q = 1 - p = 0.7

a) P(exactly three successes) = P(X = 3)

= 8C3 x 0.33 x 0.75

= 0.2541

b) P(less than three successes) = P(X = 0) + P(X = 1) + P(X = 2)

= 0.78 + 8x0.3x0.77 + 8C2x0.32x0.76

= 0.0576 + 0.1977 + 0.2965

= 0.5518

c) P(6 or more successes) = P(X = 6) + P(X = 7) + P(X = 8)

= 8C6x0.36x0.72 + 8x0.37x0.7 + 0.38

= 0.0100 + 0.0012 + 0.0001

= 0.0113

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