Originally Posted by

**lunaj** Ill post the whole thing:

**Question[/B****]: The advertised weight of a certain candy bare is 2.13 oz. however, the actual weights of these candy ars vary according to a normal distribution with mean 2.20oz and a standard deviation of 0.04 oz.**

**1)** What is the probablility that an individual candy bar weighs between 2.18 and 2.22 oz?

**2) **Suppose you plan to take a random sample of 5 candy bars and calculate the sample mean weight. Does the clt apply in this instance?

**3)** What does the clt say about how these sample means will vary from sample to sample?

**4) **Calculate the probability that the sample mean of these 5 candy bars falls between 2.18 and 2.22 oz.

**5) **How do you expect this probability to change if the sample size were 40 insetead of 5? calculate the probability that a sample mean of a random sample of 40 candy bars falls between 2.18 and 2.22.

**6)** what if the origional population had been skweed instead of normal. Which of the three probablities that you calculated ( frow q's 1 , 4, 5) remains approx. correct?

*My Response. *

**1) In the book there was a problem about babies and the probability that they weight between 3000 and 4000 grams at birth so i followed that model and answers to try and find answers for number 1.**

**ex) 2.18-2.13/0.04**

**2.22-2.13/0.04 Look it up in the table and subtract the scores and i i got .0934 so about 9%**

Mr F says: I have already said (several times) that the mean is 2.20 NOT 2.13.

**2) **No because the sample size is less than thirty and it only applies to sample thirty or greater.

Mr F says: Since the candy bars come from a normal population the CLT does apply - the distribution of the sample mean will be normal.

**3) Im not really sure but i assume that they will be very varied because there isnt many samples??**

Mr F says: Did you read the link I gave?

**4) **Im not really sure about this. When i look at it, doing this calculation would be the exact same as question one because they would be the same numbers in teh same places.

Mr F says: No. The sample mean follows a normal distribution with mean and standard deviation given by the formulae in the link I gave you. The mean is 2.20 but the standard deviation is NOT 0.04, it's $\displaystyle {\color{red}\frac{0.04}{\sqrt{5}}}$ (look at the formula in the link). So it's different to Q1.

**5)** Im not sure where 40 would go or what i would do with it.

Mr F says: Did you read the link I gave you? The sample mean follows a normal distribution - the mean is 2.20 and the standard deviation is$\displaystyle {\color{red}\frac{0.04}{\sqrt{40}}}$ (look at the formula in the link).

[b]**6) **?? It would mean that most candy bars weighed more that 2.13oz?

Mr F says: If the population was not normal, then distribution of the sample mean will only be approximately normal for large n ..... the formulae in the link I gave you will only be valid for large n.

Obviously the calculation in Q1 is no longer valid (because the population is no longer normal), the sample size in Q2 is small, the sample size in Q5 is large .....