Question

About 60% of U.S full-time college students drank alcohol within a one-month period. You randomly select six U.S. full-time students. Find the probability that the number of U.S. full-time college students who drank alcohol within one-month period isa.Exactly twob.At least threec.Less than fourd.Assume, we sampled 5000 students. What expected number of U.S. full-time drank alcohol within a one-month period?e.From part (d), find the variance and the standard deviation.

Answer 1

Here we have proportion of U.S. full time college students= p = 0.60, Random sample = n = 6

Here we apply Binomial distribution .

p ( X =x ) = ⁿC_x * p^x * ( 1 -p) ^n-x

a) Exactly two.

P ( x = 2 ) =  ⁶C₂ * 0.60² * ( 1 -0.60) ^6-2

= 0.1382

b) At least three

p ( x 3 ) = 1 - p ( x < 3 ) = 1 -{ p ( x =0 ) + p ( x =1 ) + p( x =2 ) }

= 1 - {  ⁶C₀ * 0.60⁰ * ( 1 -0.60) 6-0 +  ⁶C₁ * 0.60¹ * ( 1 -0.60) ^6-1   +  ⁶C₂ * 0.60² * ( 1 -0.60) ^6-2 }

= 1 - { 0.0041 + 0.0369 + 0.1382 }

= 1 - 0.1792

= 0.8208

c) Less than 4.

p ( x < 4 ) = p ( x =0) + p ( x =1 ) + p ( x =2 ) + p ( x =3 )

=  ⁶C₀ * 0.60⁰ * ( 1 -0.60) ^6-0 +  ⁶C₁ * 0.60¹ * ( 1 -0.60) ^6-1   +  ⁶C₂ * 0.60² * ( 1 -0.60) ^6-2 + ⁶C₃ * 0.60³ * ( 1 -0.60) ^6-3

= 0.0041 + 0.0369 + 0.1382 + 0.2765

= 0.4557

d) Here we have n = 5000

Expected number of U.S. full-time drank alcohol within a one-month period is given by,

E ( x ) = n* p = 5000* 0.60 = 3000

e) Variance :

v ( x ) = n * p * ( 1 -P ) = 5000*0.60*0.40 = 1200

Standard deviation =