Eighty four percent of workers aged 16 and over drive to wor

Eighty four percent of workers aged 16 and over drive to work alone.   For a random sample of 100 Americans,  

a.) What is the probability that exactly 90 drive to work alone?

b.) What is the probability that at most 80 drive to work alone?

c.) What is the probability that more than 78 drive to work alone?  

d.) What is the mean value of the number that drive to work alone?

e.) What is the standard deviation?  

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

a.
P( X = 90 ) = ( 100 90 ) * ( 0.84^90) * ( 1 - 0.84 )^10
= 0.0292

b.
P( X < = 80) = USE THE FORMULA IN EXCEL = ROUND(BINOMDIST(80,100,0.84,TRUE),4)
= 0.1689

c.
P( X > 78) = 1 - P ( X <=78)
P ( X <=78) = USE THE FORMULA IN EXCEL = ROUND(BINOMDIST(78,100,0.84,TRUE),4) = 0.071
P( X > 78) = 1 -0.071 = 0.929

d.
Mean = np = 100 * 0.84 = 84

e.
Standard Deviation ( npq )= 100*0.84*0.16 = 3.6661