The probability is 045 that a traffic fatality involves an i

The probability is 0.45 that a traffic fatality involves an intoxicated or alcohol-impaired driver or nonoccupant. In ten traffic fatalities, fine the probability that the number, Y, which involve an intoxicated or alcohol-impaired driver or nonoccupant is

a.

exactly three; =

at least three; =

at most three =
b. between two and four, inclusive
c. find and interpet the mean of the random variable Y
d. obtain the stand deviation of y

Solution

This is an example of binomial distribution as there are only two possible outcomes

p = probability of intoxicated driver = 0.45

q = 1 - p = probability of non-intoxicated driver = 0.55

The probability distribution of Y will bw of the form

P ( X = r ) = ⁿC_r p^r q^n-r where n = total number of observation

The probability of all the possible values of X is given below:

a)

P ( Y = 3) = 0.1664

P ( Y >= 3 ) = 1 - [ P ( Y=0) + P ( Y=1 ) + P ( Y=2) ]

= 0.9004

P ( Y < = 3) = [ P ( Y=0) + P ( Y=1 ) + P ( Y=2) + P(Y=3) ]

= 0.2660

b)

P ( 2 <= X <= 4 ) = [ P ( Y=4) + P ( Y=3 ) + P ( Y=2) ]

= 0.4811

c) mean of the distribution = 4.5

i.e in every 10 cases about 4 or 5 drivers would be alcohol impaired.

d)

Std.deviation of a binomial distribution

= sqrt (npq)

= sqrt ( 10 * 0.45 * 0.55)

= 1.5732

hope this helps