(I\'m doing this in excel) # of worms 0, 1, 2, 3, 4 probability 0.3, 0.3, 0.2, 0
ID: 3205393 • Letter: #
Question
(I'm doing this in excel)
# of worms 0, 1, 2, 3, 4
probability 0.3, 0.3, 0.2, 0.1, 0.1
I have filled a truck with 1000 randomly chosen apples.
(a) About how many apples in the truck should I expect to have no worms?
(b) I will only be paid for the truckload of apples if at least one third of them have no worms. Using the Normal approximation, what is the probability that this happens?
(c) If I want to make it more likely that more than one third of the apples have no worms, should I add apples or remove apples from the truck? (Any apples added or removed are randomly chosen without looking at if they have worms.)
Explanation / Answer
a) as probabilty of 0 worms =0.3
hence expected number of apple without worm =mean =np=0.3*1000=300
b) std deviation =(np(1-p))1/2 =(1000*0.3*0.7)1/2 =14.49
therefore probabilty that at least one third of them have no worms =P(X>=334)=1-P(X<333.5)
=1-P(Z<(333.5-300)/14.49) =1-P(Z<2.3117) =1-0.9896 =0.0104
c) we need to decrease the number of apples to make it more likely that more than one third of the apples have no worms.
As when the sample size increases, from law of large numbers, probabilty tends to reach its true mean , which in this case is 0.3.
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