Reply With Quotefirst question, you got it right.
second question:
P(X<20)= Sum (200Ci) (.04)^i (.96)^(200-i)... where i=0, sum i to 19.
Each patient that receives a particular drug has a 4% chance of dying because of the medication. If 200 patients receive this drug, what is the probability that all the patients will survive?
.96 ^ (200) = .0285% (seems pretty simple, but am I even going about this correctly?)
The next part of the question reads: What is the probability that less than 20 patients will die?
I would expect that about 8 patients (.04 * 200) would die, so I would think that there would be a rather high probability that less than 20 die. This is just common reasoning, though. Statistically, I don't know how I would go about doing this. I'm guessing a simple formula? Any guidance would be appreciated.
first question, you got it right.
second question:
P(X<20)= Sum (200Ci) (.04)^i (.96)^(200-i)... where i=0, sum i to 19.
The heights of Christmas trees grown by a nursery have a normal distribution with m = 7.0 ft and s = 0.6 ft.
a) What is the probability that a randomly cut tree will have a height less than 6.0 ft. ?
One hundred trees are cut down and sent to a local charity.
b) What is the probability that the average height of the trees cut by the nursery is between
6.90 and 7.06 ft. ?
P(X<20)= Sum (200Ci) (.04)^i (.96)^(200-i)... where i=0, sum i to 19.
thanks for the help, but one more question...
what does the Ci stand for in the equation above?
nCr means the combination of n items taking r at a time.
nCr=n!/(r! * (n-r)!) where n!=1*2*3*...n.
That is 5C2=5!/(2!*3!) Similarly 200Ci can be foundout.
But I think it is better to do this problem using Poisson distribution, taking lambda = 8.
I do not need the answers to these questions, i just need to know how to solve them!!! please!!
question 1:
A board consisting of 8 members, 3 men and 5 women must vote for or against something. If the vote is 5-3 and the members were not biased by gender, what is the probability that the vote would be split along gender lines?
question 2:
a monkey is given 12 blocks; 3 shaped like squares, 3 shaped like triangles, 3 like circles, and 3 like rectangles. It draws 3 of each kind in order, say, 3 triangles then 3 squares..etc. What is the probability of this occuring?
hello,
if you have a new question please put it in a new thread.
to the original question: i think its best solved as a binomial probability, with number of trials 200, p=.04, and number of successes 19. if you have a tool to compute the cumulative probability then you can put in the above values and be all set. if not the you will need to compute the probabilities from zero deaths up to 19 deaths and add them together.
juicefang: both of these are solved by counting the number of ways to succeed and dividing by the total number of possible outcomes.
for #1 then there is only one way in which the vote could split along gender, and the total munber of ways the vote could go can be found using the multiplication rule for counting.
cheers
jerry
If the random variable X is the no. of deaths among 200 patients, then p=.04 because ' there is a 4% chance of death ' ,according to the question. Here probability of success is 0.04, not 0.96 and the required probability in the second part of the question is P[X<20], which can be calculated either using binomial distribution with n=200 and p=0.04 or using Poisson with lambda=8. Both will give approximate same answer because n is fairly large and p is small, but the later can be more easily calculated if a tool to compute cumilative frequencies not available.
vin,
thanks for noticing my error, i corrected my post so it would not be confusing.
cheers
jerry
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