Question

The probability is 0.35 that a traffic fatality involves an intoxicated or​ alcohol-impaired driver or nonoccupant. In nine traffic​ fatalities, find 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 interpret the mean of the random variable Y.

d. Obtain the standard deviation of Y.

Answer 1

X ~ Bin( n , p)

Where n = 9 , p = 0.35

Binomial probability distribution is

P(X) = nCx px ( 1 - p)n-x

a)

P(X = 3) = 9C3 * 0.353 * ( 1 - 0.35)6

= 0.2716

P(X >= 3) = 1 - P(X <= 2)

= 1 - BINOM.DIST ( 2 , 9 , 0.35 , TRUE) (Using EXCEL )

= 1 - 0.3373

= 0.6627

P(X <= 3) = BINOM.DIST ( 3 , 9 , 0.35 , TRUE) (Using EXCEL )

= 0.6089

b)

P(2 <= X <= 4) = P(X = 2) + P(X = 3) + P(X = 4)

= 9C2 * 0.352 * ( 1 - 0.35)7 +9C3 * 0.353 * ( 1 - 0.35)6 +9C4 * 0.354 * ( 1 - 0.35)5

= 0.7072

c)

Mean = n p = 9 * 0.35 = 3.15

Interpretation = The mean number of traffic fatality involves an intoxicated or​ alcohol-impaired

driver or nonoccupant is 3.15

d)

Standard deviation = sqrt ( n p ( 1 - p) )

= sqrt ( 9 * 0.35 ( 1 - 0.35) )

= 1.4309