Question

In: Statistics and Probability

If is the fact that 13.7% of American adults smoke cigarettes. If 25 American adults were...

If is the fact that 13.7% of American adults smoke cigarettes.

If 25 American adults were chosen at random, what is the probability that at most 7 of them smoked cigarettes?

Expert Solution

Solution:

Given,

p = 13.7% = 0.137

1 - p = 1 - 0.137 = 0.863

n = 25

X follows the Binomial(25 , 0.137)

Using binomial probability formula ,

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x} ; x = 0 ,1 , 2 , ....., n

Now ,

P[at most 7]

= P[X 7]

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

= (₂₅ C ₀) * 0.137⁰ * (0.863)^25-0 + (₂₅ C ₁) * 0.137¹ * (0.863)^25-1 + (₂₅ C ₂) * 0.137² * (0.863)^25-2 + (₂₅ C ₃) * 0.137³ * (0.863)^25-3 + (₂₅ C ₄) * 0.137⁴ * (0.863)^25-4 + (₂₅ C ₅) * 0.137⁵ * (0.863)^25-5 + (₂₅ C ₆) * 0.137⁶ * (0.863)^25-6 + (₂₅ C ₇) * 0.137⁷ * (0.863)^25-7

= 0.02513447932+0.09975155465+0.19002497781+0.23127412324+0.20192937643+0.13463518331+0.07124380114+0.03069816493

= 0.98469166083

P[at most 7 ] = 0.98469166083

orchestra answered 1 year ago

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