need help A.S.A.P with these stat problems

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snm26

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1) Let x be a random variable representing the cost a childcare center charges for one child for a day. Consider a sampling distribution of sample means,(x.

a) As the sample size becomes increasingly large, what does the distribution of sample means approach?
b) What value will the mean ((x approach?
c) What value will the standard deviation ((x approach?
d) How do the (x distribution for samples of size n = 40 and n = 80 compare?


2) The number of boxes of girl scout cookies sold by each of the girl scouts in a Midwestern city has a distribution which is approximately normal with mean ( = 75 boxes and standard deviation ( = 30 boxes.

a) Find the probability that a scout chosen at random sold between 60 and 90 boxes of cookies.
b) Find the probability that the sample mean number of boxes of cookies sold by a random sample of 36 scouts is between 60 and 90 boxes?


3) Statistical Abstracts (117th edition) reports sale price of unleaded gasoline (in cents per gallon) at the refinery. The distribution is mound-shaped with mean ( = 80.04 cents per gallon and standard deviation, ( = 4.74 cents per gallon.

a) Are we likely to get good results if we use the normal distribution to approximate the distribution of sample means for samples of size 9? Explain.
b) Find the probability that for a random sample of size 9, the sample mean price will be between 79 and 82 cents per gallon.
c) Find the probability that for a random sample of size 36, the sample mean price will be between 79 and 82 cents per gallon.
d) Compare your answers for parts (b) and (c) and give a reason for the difference.


4) More than 200 billion grocery coupons are distributed each year for discounts exceeding $84 billion. However, according to a report in USA Today, only 3.2% of the coupons are redeemed. If a company distributes 5000 coupons, what is the probability that:

a) at least 100 coupons are redeemed?
b) at most 200 coupons are redeemed?
c) fewer than 100 coupons are not redeemed?
d) more than 200 coupons are not redeemed?
 
I always wonder, when it says something like this, "need help A.S.A.P", if you are sitting in an exam RIGHT NOW!! Good luck.
 
Whether or not you are sitting immediately for an exam really isn't the point. Sometimes I hope for more useful headings, that's all. However, I ALWAYS hope for some effort from the student. Could you show some, please?
 
snm26 said:
The number of boxes of girl scout cookies sold by each of the girl scouts in a Midwestern city has a distribution which is approximately normal with mean ( = 75 boxes and standard deviation ( = 30 boxes.

a) Find the probability that a scout chosen at random sold between 60 and 90 boxes of cookies.
b) Find the probability that the sample mean number of boxes of cookies sold by a random sample of 36 scouts is between 60 and 90 boxes?
Click to expand...
For part b) I think the question is miscopied.
The mean number of cookies 36 scouts will sell = [imath]75 \times 36 = 2700[/imath]
The standard deviation = [imath]\sqrt {36 \times 30^2} = 180[/imath]
Correct?
 
snm26 said:
Statistical Abstracts (117th edition) reports sale price of unleaded gasoline (in cents per gallon) at the refinery. The distribution is mound-shaped with mean ( = 80.04 cents per gallon and standard deviation, ( = 4.74 cents per gallon.

a) Are we likely to get good results if we use the normal distribution to approximate the distribution of sample means for samples of size 9? Explain.
b) Find the probability that for a random sample of size 9, the sample mean price will be between 79 and 82 cents per gallon.
c) Find the probability that for a random sample of size 36, the sample mean price will be between 79 and 82 cents per gallon.
d) Compare your answers for parts (b) and (c) and give a reason for the difference.
Click to expand...
To answer d), first the probability in c) > the probability in b) and the reason is standard deviation for c) < the standard deviation for b). How would we describe what's going on here? Have we increased our confidence that the sample mean is between 79 and 82?

Come to think of it how do we answer questions b) and c)?

I did it like this:
[imath]P\left(\frac{79 - 80.4}{\frac{4.74}{\sqrt 9}} < z < \frac{82 - 80.4}{\frac{4.74}{\sqrt 9}}\right)[/imath]

and

[imath]P\left(\frac{79 - 80.4}{\frac{4.74}{\sqrt {36}}} < z < \frac{82 - 80.4}{\frac{4.74}{\sqrt{36}}}\right)[/imath]

Is this correct?
 
Last edited:
Agent Smith said:
For part b) I think the question is miscopied.
The mean number of cookies 36 scouts will sell = [imath]75 \times 36 = 2700[/imath]
The standard deviation = [imath]\sqrt {36 \times 30^2} = 180[/imath]
Correct?
Click to expand...
No. Read the problem carefully:
snm26 said:
2) The number of boxes of girl scout cookies sold by each of the girl scouts in a Midwestern city has a distribution which is approximately normal with mean ( = 75 boxes and standard deviation ( = 30 boxes.

a) Find the probability that a scout chosen at random sold between 60 and 90 boxes of cookies.
b) Find the probability that the sample mean number of boxes of cookies sold by a random sample of 36 scouts is between 60 and 90 boxes?
Click to expand...
It isn't talking about the total number of boxes, but the mean of the number sold by each scout in a sample of 36. The mean of this sample mean will be 75.
 
Agent Smith said:
To answer d), first the probability in c) > the probability in b) and the reason is standard deviation for c) < the standard deviation for b). How would we describe what's going on here? Have we increased our confidence that the sample mean is between 79 and 82?

Come to think of it how do we answer questions b) and c)?

I did it like this:
[imath]P\left(\frac{79 - 80.4}{\frac{4.74}{\sqrt 9}} < z < \frac{82 - 80.4}{\frac{4.74}{\sqrt 9}}\right)[/imath]

and

[imath]P\left(\frac{79 - 80.4}{\frac{4.74}{\sqrt {36}}} < z < \frac{82 - 80.4}{\frac{4.74}{\sqrt{36}}}\right)[/imath]

Is this correct?
Click to expand...
@Dr.Peterson , what about ☝️
 
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