Central Limit Theorem and probability regarding xbar

[flora33]

for a random variable that is normally distributed, with mean = 80 and standard deviation = 10, determine the probability that a simple random sample of 25 items will have a mean that is:

greater than 210 = 210 – 200 / 10 = 1.000 = 0.8413; P(xbar < 210) = P(z < 1.0) = .8413
between 190 and 230 = 190 – 200 / 10 = -1 and 230 – 200 / 10 = 3; P(190 ? xbar ? 230) P(-1.0 ? z ? 3.0) = .9987 - .1587 = .84
less than 225 = 225 – 200 / 10 = 2.5; P(xbar < 225) = P(z < 2.5) = .9938

Would anyone mind confirming I have done this correctly? Thank you!

Flora

 

[royhaas]

Your calculations refer to a mean of 200 and s.d. of 10 and sample size of 1. So your calculations are correct for these parameters. However, your problem statement is different.