Am I using these formulas properly? normal distribution+CLT

[John421]

I'm attempting to work out when to use the different formulas and how everything fits together, can you confirm if the following is correct?

1) If we had the problem: There are 250 dogs at a dog show who weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 1 dog is chosen at random, what is the probability they will have a weight of greater than 8 pounds and less than 25 pounds?

Then we would use the formula: Z = (x-μ)/σ and use the z tables to work out our answer.

2) If we had the problem: There are 250 dogs at a dog show who weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 4 dogs are chosen at random, what is the probability they will have an average weight of greater than 8 pounds and less than 25 pounds?

Then we would use the formula: Z = (x-μ)/(σ/√4) and use the z tables to work out our answer.
This is the same formula as the previous formula in 1), it's just that the squareroot of 1 is 1, so Z = (x-μ)/(σ/√1) became Z = (x-μ)/σ

3) If we had the problem: There are 29 dogs at a dog show who weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 1 dog was chosen at random, what is the probability they will have an average weight of greater than 8 pounds and less than 25 pounds?

Then we would use the formula: tn-1 = (x-μ)/(σn-1/√1) = (x-μ)/(σn-1) and use the t tables to work out our answer.

4) If we had the problem: There are 29 dogs at a dog show who weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 4 dogs are chosen at random, what is the probability they will have an average weight of greater than 8 pounds and less than 25 pounds?

Then we would use the formula: tn-1 = (x-μ)/(σn-1/√4) and use the t tables to work out our answer.

Is all of this correct?

 

[mfb]

They are all good approximations if you assume normal distributions for dog weights (a bit odd, because then many dogs have negative weights).
Problem (4) can get tricky if you get more dogs for the average. In the extreme case, consider chosing 29 dogs "at random": the average will be exactly 12 pounds with probability 1, so the formula certainly breaks down.