On average, there are three misprints in every 10 pages of a particular book. If every chappter of the book contains 35 pages, what is the probability that chapters 1 and 5 have 10 misprints?
Just would like to know if I am right or not. If not could you point me in the right direction?
I think this is the answer to solve it by binomial distribution and here what I got
N being total pages and P being percent of occurrences ( 3/10 = .3 )
n = 35 p = .3
Normal distribution just those two types of distribution. I thought binomial because the distribution is yes/no there either is mistake or their isn't one. Should I have used normal distribution but that doesn't seem to be the case to me am I wrong?
The distribution is Poisson as the probability of occurance of success (i.e. the misprints) is too low as compared to the total number of cases (only 3 misprints out of 10 pages). here lamda=4.28 [(3*10*5)/35],x=10...hence on counting you will get the answer...
If lambda 18.3428571 and x = 10 (being the occurances) then 0.0128 is the answer I got
Equation being f(x, lambda) = [(lambda^x)(e^lambda)]/ x! using tablet calculator got me 0.0128 with the lamba of 4.28[(3*10*5)/35]
Doesn't that seem a little low considering if there 3 mistakes every 10 pages and you have 35 pages you'd almost think there would be between 9,10,11 . I think a higher percent then .7%
sorry i did a small mistake. 3 misprints per 10 pages = 3/10 misprints per page. now one chapter has 35 pages means (3*35)/10 misprints. now there are 5 chapters....so total misprints= (3*35*5)/10 misprints = 52.5. this 52.5 is the lamda i think.
So if λ is ( .3 * 35 ) you get 10.5 plug that with Prob being exactly 10 you get .1236 Poisson Distribution and Cumulative Poisson Distribution of .521 . Don't think the cumulative one matters but just incase
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