Math Probability (hypergeometric distribution) questioin (# 67)?

Question

a geologist collected 10 specimen of basaltic rock and 10 specimens of granite.an assistant randomly selects 15 of the specimen for analysis.

a) what's the pmf of the # of granite specimes?

b) what's the probability that all specimens of one of the two types of rock are selected for analysis?

c) what is the probability that the number of granite specimens selected for analysis is within 1 devistation of its mean value.




here're the answers from the back of the book
a) h(x,15,10,20) x=5...10
b) .0325
c) .697


My question is how to approach and solve part C only

thanks a bunch.

lol. sry it should be deviation.

BUT. where are you guys picking up 7 and 8? why not 9 and 10?

Sry. Can you clarify how to achieve the standard deviation?

Answer 1

for this hypergeometric distribution, the mean of X [X = the number of granite specimens selected] is 15 x ‎½ = 7.5.

using the formula for the variance of the hypergeometric distribution, the variance of X is 20(20 - 15) x ‎½ x ‎½ /15 =‎ 5/3 = 1.67, so the standard deviation is ‎√(5/3) = 1.3.

so to be within one sd of the mean, X has to be between 7.5 - 1.3 = 6.2 and 7.5 + 1.3 = 8.8. since X is an integer, X must be 7 or 8.

then the hypergeometric distribution gives

P(X = 7) = C(10,7) x C(10,8) / C(20,15) = .3482.

[here C(10,7) means 10 choose 7 or 10!/7!3! = 120.]

similarly P(X = 8) = .3482. [to see this quickly, note that if X = 8, then Y = number of basaltic rocks selected is 7. that is, the events (X = 8) and (Y = 7) are the same.

by symmetry, X and Y have the same probability distribution, so

P(X = 8) = P(Y = 7) = P(X = 7) = .3482.]

then P(X = 7 or 8) = 2x.3482 = .6964.

Answer 2

Ok, to solve part c, you need to know part a.

First, we need to know the mean of a hypergeometric distribution. You can check out any statistic book for the proof, but I won't give the proof here.

The mean is nD/N which in this case n = 15, D = 10 and N = 10 + 10 = 20.

So the mean is (15*10)/20 = 7.5

So far okay?

Now the deviation of 1 from the mean value, think about it, only when the number of granite specimens being 7 or 8 will fit this criteria. Any other number will fail and you certainly can't have 6.5 or 8.5 granite specimens. Any number not 7 or 8 will have more than 1 deviation from the mean which is 7.5
For example 9 deviates from 7.5 by 1.5 and 10 deviates from 7.5 by 2.5, both 2.5 and 1.5 are greater than 1, which fails the criteria u were given.

Let k be the number of granite specimens,

So, all one has to do is to find P(k=7) + P(k=8) using the pmf givein in (a).

P(k=7) = (10C7 * 10C8)/(20C15)
P(k=8) = (10C8 * 10C7)/(20C15)

add them together and you will get 0.697 as the answer approximated to 3 dec places.

I get 0.69659442724 on my calculator.

Hope this helps.

Answer 3

assuming you are asking about one std. deviation from the mean

If u is the mean and s the standard deviation,
u = 7.5, s = 0.993

reqd. prob = P{ |X- u| < s }

= P{-s < X-u < s}

= P{ u-s < X < u+s}

substitute the values for mean u and std. deviation s and evaluate the probability.

reqd. prob = P{ 6.51 < X < 8.49}

= P(X=7) +P(X=8)

=0.697(approximately)