Sampling Distrubtions

Question

On the distant planet Cowabunga, the weights of cows have a normal distribution with a mean of 597 pounds and a standard deviation of 75 pounds. The cow transport truck holds 5 cows and can hold a maximum weight of 3275. If 5 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 3275?

Give answer correct to 4 decimal places.

Answer 1

Cowabunga! That's a lot of Bull. short answer is .02936.

Long Answer.

Well, let us see. Mean size of a walking hamburger is 597 with a standard deviation of 75 pounds. Since the question is not about the mean of a sample but the size of a sub population we don't have to find out the standard deviation of the mean. BTW - That parameter (standard error) is totally useless in this problem. This makes things really simple.

note: the math behind the variance of a kine of cattle can easily be demonstrated but is eliminated for clarity.

Since the weight of each cow is independent of the others and identical, it can be shown that the variance of the sum of the 5 cows is equal to the sum of the variance of the individual cows and the mean of the kine is the sum of the mean of each cow. Thus

V_total= 5 * (σ_cow)² or 28125 and the standard deviation of the kine (what we call a group of less than 200 cows) or σ_kine is 167.70 lbs & μ_kine = 2958 lbs

Now comes the statistics. We have a mean for the kine (ok, call it a group if you will , but not a herd) of 2958

We have a standard deviation for the kine of 167.70. Find the probability that we will overload the truck.

Using a TI-8x model calculator with statistics functions)

P(w>3275) is nmcdf(3275,9e9,2958,167.70) = .02936. (rounded this is 0.0294)

NOTE: This is NOT the same as saying that the average weight of a kine of 5 cows would have a probability of .0294 of being larger than 655. Don't calculate that, it would lead to the wrong answer. (P(x>655)=.2196) Please don't do this. The distribution of the mean weight of a kine is NOT the distribution of the kine itself.