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

Step-by-step explanation:a.-

a-Notice that this is a binomial probability distribution problem that is generally expressed as:

Given that p=0.45, n=10, the probability of exactly 3 intoxicated drivers is:

Hence, the probability of exactly  three intoxicated drivers  is 0.1665

ii. The probability of at least 3 intoxicated drivers:

We use the value of p=0.45 and n=10 and (1-p)=0.55:

Hence, the probability off at least three intoxicated drivers is 0.9005

iii. The probability of at most three intoxicated drivers:

-This is calculated the probabilities of between 0 to 3 as:

Hence, the probability off at most three intoxicated drivers 0.0995

b. The probability of between tow and four, inclusive, is calculated by summing the exact probabilities of x=2,3 and x=4:

Hence, the probability of between 2 and 4 is 0.4812

c. The mean of the binomial random variable is calculated as:

Hence, the mean of random variable Y is 4.5

d.Given that n=10 and p=0.45, the standard deviation of the binomial random variable Y is calculated as:

Hence, the standard deviation of the random variable Y is 1.5732

nada

step-by-step explanation:

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