NEED HELP for STATS - don't know where to start!!!

Question

2. You have designed an experiment to investigate the behavior of Himalayan hamsters. In this experiment, a hamster is shown five closed boxes. Two of these boxes contain treats while the others are empty. The hamster is encouraged to select 3 boxes, one at a time, and to see if the box it has selected contains a treat. Each time the hamster selects a box, the box is opened and then removed so that the hamster cannot select it again.

a) Generate the probability distribution of the variable X: the number of times a hamster selects a box containing a treat.

b) Express the maximum number of boxes containing a treat that a hamster could select as a Z score.

Answer 1

This distribution is an example of a hypergeometric distribution, when we count members of a subpopulation in a sample drawn without replacement, e.g., the distribution of spades in a five card poker hand. Despite the name, this is more similar to a binomial distribution than a geometric distribution. It is a little more concentrated around the mean.

There are 5 choose 3 = 5!/(3!2!) = 10 ways to choose 3 boxes out of 5. You can write these out to check. The number of ways to choose 3 empty boxes and 0 full boxes is (3 choose 3)(2 choose 0) = 1. The number of ways to choose 2 empty boxes and 1 full box is (3 choose 2)(2 choose 1) = 6. The number of ways to choose 1 empty box and 2 full boxes is (3 choose 1)(2 choose 2) = 3. We divide these counts by the total number of equally likely outcomes to get probabilities of 1/10, 6/10, and 3/10 that X = 0, 1, and 2 respectively.

Now that you have the distribution, you can calculate the mean and the population standard deviation. The Z-score of the raw maximum 2 is (2-mean)/(standard deviation), the number of standard deviations above the mean.

Answer 2

0, 1 or 2 treats possible in random selections

either no treat 3 times in a row, only 1 way to do that

or a treat on the 1st, 2nd or 3rd try, 3 ways to do that

or no treat on the 1st 2nd or 3rd try, 3 ways to do that

7 possible outcomes

P(0)= 1/7

P(1) = 3/7

P(2) = 3/7

P(x)= 1/7 if x =0, and 3/7 if x=1 or 2

Pr(0)+Pr(1)+Pr(2) = 1/7 +3/7 +3/7 = 7/7 = 1

Pr(x) =3/7 unless x=0 then P(x)=1/7

draw a tree diagram

TTN

TNT

TNN

NTN

NTT

NNT

NNN

7 outcomes, 1 has 0 treats, 3 have 1 treat, 3 have 2 treats

Pr(0)=1/7, Pr(1)=3/7=Pr(2)

f(x) = -(x-3/2)^2 +13/4, a downward opening parabolic function going through (0,1), (2,3), (1,3)

of f(x) = sqr(13/4-x)+3/2 for the points (1,0), (3,2), (3,1)

or

Pr(x) = (-1/7)(x-3/2)^2 +13/28 which gives probability of x = 0, 1 or 2

mean number of treats = (1(0) + 3(1) +3(2))/7 = 9/7 = 1 2/7

mean has z score = 0, E(x)= Expected # of treats = 9/7

calculate standard deviation to find the other z scores

max X = 2

z for 2 = (2- 9/7)/sd

sd = sqrVar(x)

Var(x)=sum of squared differences/n

2-9/7 = 5/7 (5/7)^2 = 25/49

1-9/7 =-2/7, (-2/7)^2 = 4/49

0-9/7 = -9/7 (-9/7)^2 = 81/49

sum of squared differences = 110/49

110/49/7 = 110/7= about 15.7 (treating this as a population, some might divide by 6 if they treat it as a sample)

sd = sqr(110/7) = almost 4 = about 3.96

z score = about .18 for x=2, (2-9/7)/3.96= about .18