Question

A report by the Traffic Injury Research Foundation stated that alcohol was a factor in 37 % of accidents causing the fatally injury of automobile drivers. In a random sample of 12 automobile accidents resulting in the death of the driver:

a) What is the probability that alcohol was not a factor in any of the accidents?

b) What is the probability that alcohol was a factor in more than 2 but less than 7 of the accidents?

c) What is the expected number and standard deviation of accidents in which alcohol was a factor?

Answer 1

we have p = 0.37 and n(sample size) = 12

(a)  probability that alcohol was not a factor in any of the accidents

Using binomial formula, we have total 12 autmobile and we have to select 0 because we are considering that alcohol was not a factor

P(x =r) = C(n,r)*(p^r)*(q)^(n-r)

putting r = 0 and n = 12, we get

P(x =0) = C(12,0)*(0.37^0)*(1-0.37)^(12-0) = 12!/[(12-0)!*0!]*0.37^0*(0.63^12) = 0.0039

so, the  probability that alcohol was not a factor in any of the accidents is 0.0039

(b) probability that alcohol was a factor in more than 2 but less than 7 of the accidents means alcohol was a factor for 3,4,5 and 6 of the accidents

using the same formula which is stated in above part, we can write

P(x =3) = C(12,3)*(0.37^3)*(1-0.37)^(12-3) = 12!/[(12-3)!*3!]*0.37^3*(0.63^9) = 0.1742

P(x =4) = C(12,4)*(0.37^4)*(1-0.37)^(12-4) = 12!/[(12-4)!*4!]*0.37^4*(0.63^8) = 0.2302

P(x =5) = C(12,5)*(0.37^5)*(1-0.37)^(12-5) = 12!/[(12-5)!*5!]*0.37^5*(0.63^7) = 0.2163

P(x =6) = C(12,6)*(0.37^6)*(1-0.37)^(12-6) = 12!/[(12-6)!*6!]*0.37^6*(0.63^6) = 0.1482

adding all these probabilities, we get P( more than 2 and less than 7) = 0.1742+0.2302+0.2163+0.1482=0.7689

so, the probability that alcohol was a factor in more than 2 but less than 7 of the accidents is 0.7689

(c) Expected number = n*p

where n is sample size = 12 and p is probability = 0.37

so, expected number = 12*0.37 = 4.44 or 4 accidents (rounded to nearest whole number)

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

here sqrt means square root

value of n = 12, p = 0.37

we get

Standard deviation = sqrt(12*0.37*(1-0.37)) = 1.6725( 4 decimals) or 1.67 (2 decimals)

you can use any number of decimals based on your question requirement