normal distribution

Taya

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Ok I will give you two problems, the first problem I have already done, the reason I am showing you the first problem is because the second problem piggybacks off this one, so here is the first problem which i already got the answer to (hopefully correct) :


  1. Student scores follow a Normal Distribution with the mean of 1100 and a standard deviation of 150. What percentage of individual students would you expect to get a score of 1150 or higher? now what I got was
z=x-1100/150 z=1150-1100/150=.33 this is the answer i got now for the 2nd problem asked:

we sample 35 students with same mean and standard deviation as above. If we were to take the average of the 35 students’ scores, what is the probability that their average would be higher than 1150?

The fact that it is asking for the average throws me off. Is this problem supposed to be set up like the first one?
 
..., so here is the first problem which i already got the answer to (hopefully correct) :


  1. Student scores follow a Normal Distribution with the mean of 1100 and a standard deviation of 150. What percentage of individual students would you expect to get a score of 1150 or higher? now what I got was
z = (x - 1100)/150 \(\displaystyle \ \ \ \)z = (1150 - 1100)/150 = .33 for the rounded value
Put parentheses around the differences that are in the numerators because of the Order of Operations.
\(\displaystyle \ \ \).33 is not the answer. Look up .33 in a table of positive z-scores. \(\displaystyle \ \ \) .6293 is the entry, but that represents the area under the curve from the extreme left-hand side of the curve up to that point.\(\displaystyle \ \ \) So subtract .6293 from 1 to get .3707. \(\displaystyle \ \ \) Then, about 37 out of every 100 students would be expected to get a score of 1150 or higher. \(\displaystyle \ \ \ \ \)I don't know what is intended for the second question.
 
Ok I will give you two problems, the first problem I have already done, the reason I am showing you the first problem is because the second problem piggybacks off this one, so here is the first problem which i already got the answer to (hopefully correct) :


  1. Student scores follow a Normal Distribution with the mean of 1100 and a standard deviation of 150. What percentage of individual students would you expect to get a score of 1150 or higher? now what I got was
z=x-1100/150 z=1150-1100/150=.33 this is the answer i got now for the 2nd problem asked:

we sample 35 students with same mean and standard deviation as above. If we were to take the average of the 35 students’ scores, what is the probability that their average would be higher than 1150?

The fact that it is asking for the average throws me off. Is this problem supposed to be set up like the first one?
lookagain said:
.33 is not the answer. Look up .33 in a table of positive z-scores. .6293 is the entry, but that represents the area under the curve from the extreme left-hand side of the curve up to that point. So subtract .6293 from 1 to get .3707. Then, about 37 out of every 100 students would be expected to get a score of 1150 or higher.
You have a population distribution with mean 1100 and std.dev. 150, from which you take a sample of size N=35. What does the "Sampling Theorem" tell you about the distribution of sample means? Where does 1150 fall in that distribution?
 
z score

thank you soooo much! on question though am I looking at the areas under the normal curve table? thats what confuses me because i dont see where you found the .6293 number.
Appreciate the help!!
 
thank you soooo much! on question though am I looking at the areas under the normal curve table? thats what confuses me because i dont see where you found the .6293 number.
Appreciate the help!!
Yes, we are using tables of the area under the normal curve. In my table the first column is called "x," and is the same as the z-score. That is where I look up the entry for z (or x) = 0.33. The next column is the area from -infinity up to x, which is 0.6293. Another possibility for your table could be to give the area between the midpoint and x, which would be 0.1293. The area greater than x is 0.3707, or 37%.

BTW, if you interpolate for z=0.3333.. to be 1/3 of the way between 0.33 and 0.34, the more precise result for the area would be 0.3694 - still rounds to the same 37% (close enough!).

What answer have you gotten for part 2 of the question?
 
problem 2

i dont know why I cant figure this out but ive tried here it is. its like a part two question and i have already gotten the answer for the first part.

first part question is: Let’s assume that nationwide individual scores follow a Normal Distribution with the mean of 1100 and a standard deviation of 150. What percentage of individual students would you expect to get a score of 1150 or higher?


1150-1100/ 150

50/150=.33
.33= .6293
1-.6293=.3707 so 37%

then here is the second part with same mean and standard deviation as above

we sample 35 students from a school mean is 1100 and Standard deviation is 150. If we were to take the average of the 35 students’ scores, what is the probability that their average would be higher than 1150? i cant figure it out! i get the first part but finding the average is throwing me off! please help!
 
i dont know why I cant figure this out but ive tried here it is. its like a part two question and i have already gotten the answer for the first part.

first part question is: Let’s assume that nationwide individual scores follow a Normal Distribution with the mean of 1100 and a standard deviation of 150. What percentage of individual students would you expect to get a score of 1150 or higher?
Taya, again, do not leave out the grouping symbols of the difference expression that is in the numerator.

(1150 -1100)/150

z-score = 50/150 = .33 . . . . . That is correct when rounded to the hundredths place
.33 = .6293 . . . . . No, the positive z-score of .33 corresponds to the area under the bell-shaped curve (read: probability) of .6293. There is not supposed to be an equals sign between them there.
1 - .6293 = .3707 so 37% . . . . .Yes, that is correct (to the nearest whole percent).
.
 
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then here is the second part with same mean and standard deviation as above

we sample 35 students from a school mean is 1100 and Standard deviation is 150. If we were to take the average of the 35 students’ scores, what is the probability that their average would be higher than 1150? i cant figure it out! i get the first part but finding the average is throwing me off! please help!
You asked this same question before and Dr Phil responded to "You have a population distribution with mean 1100 and std.dev. 150, from which you take a sample of size N=35. What does the "Sampling Theorem" tell you about the distribution of sample means? Where does 1150 fall in that distribution?" The "sampling theorem" (there are a number of "sampling theorems" having to do with sampling time series but I am pretty sure this is the one intended!) says that if we take a sample of size n from a population with mean \(\displaystyle \mu\) and standard deviation \(\displaystyle \sigma\), the average of the sample has normal distribution with mean \(\displaystyle \mu\) and standard deviation \(\displaystyle \frac{\sigma}{\sqrt{n}}\).
 
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