At a certain plant, the maximum number of accidents occurring every day was repo

Question

At a certain plant, the maximum number of accidents occurring every day was reported to 4 Assume that the accidents occur independently every day. The probability of daily accidents is listed as below: No. of Daily Accidents Probability 0.1 0.02 0.005 0.001 a) What is the probability of at least 2 accidents will occur tomorrow given that no accidents occur today? b) What is the standard deviation of daily number of accidents? c) To control the accidents, a penalty of $20,000 will be incurred when 3 accidents occur on a day and a severe penalty of $50,000 will be incurred when 4 accidents occur on a day. How much will be expected to pay for the accidents penalty annually?

Explanation / Answer

So the actual prob table will look like as below

number of daily accidents : 0 1 2 3 4

probability : 0.874 0.1 0.02 0.005 0.001

how did I calculte 0's probability is (1-0.1-0.02-0.005-0.001)

1)Probability of atleast 2 accidents will happen tomorrow given there is no accident today = P(2)+P(3)+P(4)

since accidents occur independently every day

so., ans = 0.02+0.005+0.001=0.026

2) sd of accidents

mean = 0*0.874+1*0.1+2*0.02+3*0.005+4*0.001=0.159

sd= sqrt(0.0874*(0-0.159)^2 + 0.1 *(1-0.159)^2 + 0.02*(2-0.159)^2+0.005*(3-0.159)^2+0.001*(4-0.159)^2) = 0.4425302

So., the std deviation of accidents is 0.4425302

3) EV or expected pay = prob * penalty = 0.005*20000+0.001*50000= 150$

Hope the above explaination has helped you in understanding the problem Pls upvote the ans if it has really helped you. Good Luck!!

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