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QUESTION

Solve the following items on a clean sheet of paper.

1. The weight of goats at a farm is normally distributed with a mean of 60 kg and a standard deviation of 10 kg. A truck used to transport goats can only accommodate not more than 650 kg. If 10 goats are selected at random from the population, what is the probability that the total weight exceeds the maximum weight?
2. The blood groups of 200 people are distributed as follows: 40 have tyYpe A blood, 75 have type B blood, 60 have type O blood, and 25 have tyYpe AB blood. If a person from any of the group is selected at random, what is the probability that this person has an O blood type? How about the AB blood type?
3. The number of accidents in a production facility has a Poisson distribution with a mean of 2.8 per month. For a given month, what is the probability that there will be more than three (3) accidents?

ANSWER:

1) 801? Let X be a reandom variable denoting the weight of a goat .. X NN ( 60, 10 ) det X,, X, ... X, be the weights of to randomly selected goats. We are to find ? ( X , + X , + ... X , > 650 ) X, +x, f... Xlo. 650 10 10 = P ( x > 65.@) (x) 0) ? We know, ( 8 ) BAND In then. L J X; = X ~Ne, 5 mm) 1 = 1 (5 ) CWD In our case 80.07 N. N / 60, 10 ) : 1= 60, 17: 10 #, n = 10) - ) X NN ( GO , (TO ) " ) P ( X > 65 ) X - 60 65- 60 VIO 80.0 52 .0 = P ( Z > 1 . 58 ) 80.0 RUN= 0.5- 0.4429 = 0. 05 71Blood type No. of people A 40 75 wwp O 60 AB 25 total 20 01(0 * ) .. Probability that the person has an O blood type No . of people with blood type o" (8 total no. of people 166 200 0. 3 Similarly, probability that the person has an No of people with blood type " AB" AB blood type = total no. of people 08 08.0 200 = 0. 125 ( 3) 801." Let X be a random variable denoting the number of accidents in a production facility * ~ Poisson ( 2 . 8 ) P ( X = X ) = ( 2 . 8 ) ; x = 0 , 1 , 2 , + . ". For a given month, the probability that there will be more thein 3 accidents is O given by P ( X > 3 ) = P ( x = 4 ) + P ( x = 5 ) + . " = 1 - 1 8 ( x = 0 ) + P ( x = 1 ) + P ( x = 2 ) + P (x = 3) ] - 2.8 -2 . 8 = 1 . e (2.8) e ( 2 .8 ) - 2 .8 e ( 2.8 ) 01 2! - 2.8 ( 2 . 8 ) 31 - 2 . 8 1 - e 1 + 2.8 + 3. 92 + 3.66 1 in hoold spice Jyo 2 1 - 0. 6920 = 40 0. 3080 .. The read probability is 0.3080