1. ## Help please, need some clarification on probability!

I am a little confused about a couple problems, i think because the word mean is used. Ill write them down. any help would be greatly appreciated!

PROBLEM : The advertised weight of a certain candy bar is 2.13 oz. however the actual weight can vary to a normal distribution with a mean of 2.20 oz and stadard dev. of 0.04 oz.

1) what is the probability that an individual candy bar weighs between 2.18 and 2.22 oz.

i found the z score for both so

2.22-2.13/ .04 and 2.18-2.13/0.04 i looked in teh table 2, subtracted them from each other and got .0934 so about 9 %??

I also assume that the CLT does not apply beause the sample is less than 10.

2) what does the clt say about how these means will vary from sample to sample? I wasnt really sure, i just know that as the sample size increases it will be more "normal"

3) calculate the probability that the sample mean of theses 5 candy bars falls between 2.18 and 2.22 oz.

do i just do 2.18-2.20/0.04 ( and the same for 2.22) and subract the z scores? Similar to the first problem??

4) how do you expect the prob. to change is the sample size were 40 instead? calculate the prob that the candy bar falls between 2.18 and 2.22?(d)

I have no idea where 40 would go.

5) What if the origional population had been skewed instead of normal. which of the 3 ( Q's 3, 4 and 1) probablilities that you calculated remains approx. correct?

help!
thanks so much!

2. Originally Posted by lunaj
I am a little confused about a couple problems, i think because the word mean is used. Ill write them down. any help would be greatly appreciated!

PROBLEM : The advertised weight of a certain candy bar is 2.13 oz. however the actual weight can vary to a normal distribution with a mean of 2.20 oz and stadard dev. of 0.04 oz.

1) what is the probability that an individual candy bar weighs between 2.18 and 2.22 oz.

i found the z score for both so

2.22-2.13/ .04 and 1.18-2.13/0.4 i looked in teh table 2, subtracted them from each other and got .0934 so about 9 %??

I also assume that the CLT does not apply beause the sample is less than 10.

2) what does the clt say about how these means will vary from sample to sample? I wasnt really sure, i just know that as the sample size increases it will be more "normal"

3) calculate the probability that the sample mean of theses 5 candy bars falls between 2.18 and 2.22 oz.

do i just do 2.18-2.20/0.04 ( and the same for 2.22) and subract the z scores? Similar to the first problem??

4) how do you expect the prob. to change is the sample size were 40 instead? calculate the prob that the candy bar falls between 2.18 and 2.22?(d)

I have no idea where 40 would go.

5) What if the origional population had been skewed instead of normal. which of the 3 ( Q's 3, 4 and 1) probablilities that you calculated remains approx. correct?

help!
thanks so much!
In Q1 I have no idea where the values you've used to get the z-scores have come from. The given mean is 2.20 NOT 2.13. The smaller value of X (the weight of a candy bar) is 2.18 NOT 1.18. The sd is 0.04, not 0.4 as you have in the second z-score. Furthermore, the CLT is completely irrelevant to the required probability calculation.

As for the other questions, clearly the distribution of the sample mean is the key concept. But details about samples, sample sizes etc. are not mentioned at all in the preamble to the question .....???

The distribution of the sample mean (sampling distribution of the mean) should be extensively discussed in your class notes and textbook. You would be strongly advised to review that material.

3. sorry the first question was what is the probability that an individual candy bar weighs between 2.18 and 2.22 oz

'* sorry for the typing errors 2.18 and 0.04

Im taking an online class, the book that were using doesnt really go into detail about the sample mean, its more generalized. I tried looking but didnt find much helpful. (workshop statistics-discovery with data and fathom)

4. Originally Posted by lunaj
sorry the first question was what is the probability that an individual candy bar weighs between 2.18 and 2.22 oz

'* sorry for the typing errors 2.18 and 0.04
[snip]
I suspected as much. But the real error is that the mean of the distribution is 2.20 NOT 2.13:

"the actual weight can vary to a normal distribution with a mean of 2.20 oz and stadard dev. of 0.04 oz."

Originally Posted by lunaj
[snip]
Im taking an online class, the book that were using doesnt really go into detail about the sample mean, its more generalized. I tried looking but didnt find much helpful. (workshop statistics-discovery with data and fathom)

Originally Posted by mr fantastic
As for the other questions, clearly the distribution of the sample mean is the key concept. But details about samples, sample sizes etc. are not mentioned at all in the preamble to the question .....???
Have you posted the entire question exactly as it's written?

5. Ill post the whole thing:

[B]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%

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

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

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.

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

6) ?? It would mean that most candy bars weighed more that 2.13oz?

6. 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??

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 ${\color{red}\frac{0.04}{\sqrt{5}}}$ (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 ${\color{red}\frac{0.04}{\sqrt{40}}}$ (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 .....
You need to really review the whole topic od sampling distributions and the CLT.

Note: The information "The advertised weight of a certain candy bare is 2.13 oz." is NOT relevant to Q1 - 6.

7. thanks! this helped !