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A college surveyed 50 students to determine the average 18 Jun 2006, 06:33
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A college surveyed 50 students to determine the average number of minutes studied per night. If 84 percent of the students studied for at least 114 minutes per night and the students’ responses had a normal distribution, what is the average number of minutes studied per night?

(1) If 10 more students averaging 192 minutes per night were added to the sample, the new average would be 10 percent greater.

(2) Only 1 student studied at least 132 minutes per night

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A college surveyed 50 students to determine the average Updated on: 18 Jun 2006, 07:47
Originally posted by giddi77 on 18 Jun 2006, 06:55.
Last edited by giddi77 on 18 Jun 2006, 07:47, edited 1 time in total.
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D?
For this problem we need to know that 68% of the values in a normal distribution are with 1 sgima and 96% are within 2 sigma. I am not sure if we are expected to know this for GMAT??

Here is a link describing Normal Distribution in detail https://en.wikipedia.org/wiki/Normal_Distribution
84%
1. If x is the average
(50*x+192*10)/(50+10) = (1+10/100)*x
=> x = 120. SUFF

2. 84% of the students are above 114 => 114 is 1 sigma below the average.
Only 1 student studies 132.
Since 96% of 50 are below 2 sigma i.e, 2 are below 2 sigma. Among these 2 one value is 2 sigma above and onother is 2 sigma below.
So (2*sigma) + (1*sigma) = 132-114 = 18
3*sigma = 18
sigma = 6.

Average = 114+6 = 120. SUFF
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A college surveyed 50 students to determine the average 18 Jun 2006, 08:43
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giddi77
D?


2. 84% of the students are above 114 => 114 is 1 sigma below the average.
Only 1 student studies 132.
Since 96% of 50 are below 2 sigma i.e, 2 are below 2 sigma. Among these 2 one value is 2 sigma above and onother is 2 sigma below.
So (2*sigma) + (1*sigma) = 132-114 = 183*sigma = 18
sigma = 6.

Average = 114+6 = 120. SUFF

Giddi77,
could you explain how you got the expression marked in bold.
Show more


Check this link:
https://www-stat.stanford.edu/~naras/jsm/NormalDensity/NormalDensity.html

In this problem,
x = average
u = sigma
x-u = 114 --(a)
x+2*u = 132 --(b)

(b)-(a) gives
3*u = 18
u = 6

=>x = 114+6 = 120

Again I am not sure we need know all this stugg for GMAT but doesn't hurt to know
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A college surveyed 50 students to determine the average 21 Aug 2006, 12:30
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giddi77
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D?


2. 84% of the students are above 114 => 114 is 1 sigma below the average.
Only 1 student studies 132.
Since 96% of 50 are below 2 sigma i.e, 2 are below 2 sigma. Among these 2 one value is 2 sigma above and onother is 2 sigma below.
So (2*sigma) + (1*sigma) = 132-114 = 183*sigma = 18
sigma = 6.

Average = 114+6 = 120. SUFF

Giddi77,
could you explain how you got the expression marked in bold.

Check this link:
https://www-stat.stanford.edu/~naras/jsm/NormalDensity/NormalDensity.html

In this problem,
x = average
u = sigma
x-u = 114 --(a)
x+2*u = 132 --(b)

(b)-(a) gives
3*u = 18
u = 6

=>x = 114+6 = 120

Again I am not sure we need know all this stugg for GMAT but doesn't hurt to know
Show more


Hi.. i still don't get .. plase can u tell me.. how i arrived at satement (a )
x-u = 114 --(a)

i can 't udnersatnd why 114 is 1SD below teh mean..
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A college surveyed 50 students to determine the average 21 Aug 2006, 14:23
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shinewine

Hi.. i still don't get .. plase can u tell me.. how i arrived at satement (a )
x-u = 114 --(a)

i can 't udnersatnd why 114 is 1SD below teh mean..
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Do you know the properties of the Normal Distribution? How SD and mean aere related? How many percent of values of a set will lie in 1 SD, 2SD 3SD etc.
First read them and let me know. I don't want to spoon feed you.
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A college surveyed 50 students to determine the average 29 Oct 2006, 20:36
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I take A.

say the total is T.

from 1: T/50 * (1 + 10%) = (T + 192*10)/(50 + 10)

we can solve the total. and we can find the average by T/50.
so 1 is suff.


from 2: 1 student more than 132 min, we dont know the rest. and we don't know the average. so the normal distribution says nothing about SD either.

I can't find teh average from 2.


I hope to see other answers.
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A college surveyed 50 students to determine the average 14 Jul 2011, 13:27
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tennis_ball
I take A.

say the total is T.

from 1: T/50 * (1 + 10%) = (T + 192*10)/(50 + 10)

we can solve the total. and we can find the average by T/50.
so 1 is suff.


from 2: 1 student more than 132 min, we dont know the rest. and we don't know the average. so the normal distribution says nothing about SD either.

I can't find teh average from 2.


I hope to see other answers.
Show more

Sorry to start an old thread again. Was hoping someone could explain why statement 2 is sufficient. Here is for statement 2.

The question tells us that 84% of the students study for at least 114 minutes per night. Remember that the important standard deviation percentages are 34-14-2 ?, and that 114 is one standard deviation above the mean (100 – 2 + 14). Statement 2 tell us that only one out of fifty, or 2%, studies at least 132 minutes; thus 132 is three standard deviations above the mean. Using the two average values we know, we can calculate the difference between each average which equals the standard deviation \(\frac{(132-114)}{3}=6\). Since the mean is one standard deviation below 114, then we know that the mean is 120.
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A college surveyed 50 students to determine the average 14 Jul 2011, 22:19
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A few things here:

* properties of normal distributions are *not* tested on the GMAT. So this question is irrelevant for GMAT test takers (you don't need to know the 34-14-2 'rule', for example);

* I've seen several questions posted on gmatclub about normal distributions, all from the same source, and in almost all of them the math is simply wrong (I'd be curious to know the source of this question);

* the normal distribution is an *infinite* distribution. All of the questions I've seen posted here claim that a *finite* set is 'normally distributed'. That is a mathematical impossibility; a finite set can only be 'approximately normal'. This is the reason why most of these questions don't make any mathematical sense. The explanation below, which I gather you copied from the original source:

Chetangupta

The question tells us that 84% of the students study for at least 114 minutes per night. Remember that the important standard deviation percentages are 34-14-2 ?, and that 114 is one standard deviation above the mean (100 – 2 + 14). Statement 2 tell us that only one out of fifty, or 2%, studies at least 132 minutes; thus 132 is three standard deviations above the mean. Using the two average values we know, we can calculate the difference between each average which equals the standard deviation \(\frac{(132-114)}{3}=6\). Since the mean is one standard deviation below 114, then we know that the mean is 120.
Show more

does not make any sense - not only is it consistently using the word 'below' when it means 'above' and uses a faulty equation, but it's also logically wrong. The logical problem in the solution above can be illustrated more simply. If I ask:

Set S contains 50 temperature readings. If the mean of S is 100, what is the standard deviation of S?
1. Exactly one of the values in S is greater than 120.
2. Exactly one of the values in S is more than 2 standard deviations above the mean.

then using both statements, it's certainly possible that the standard deviation of S is 10, but that does not *need* to be true. Maybe the standard deviation is 11, and the value which is greater than 120 is equal to 125, say. The answer here is E.

The same is true in the problem above. Yes, we know that 1 of 50, or 2%, of students studied for longer than 132 minutes. That does not mean the value 132 is exactly two standard deviations above the mean, as the solution above claims. It might be that the value 131.7 is exactly two standard deviations above the mean, and that there are just no values between 131.7 and 132 in the set. So Statement 2 is certainly not sufficient here.
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A college surveyed 50 students to determine the average 15 Jul 2011, 01:36
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IanStewart
A few things here:

* properties of normal distributions are *not* tested on the GMAT. So this question is irrelevant for GMAT test takers (you don't need to know the 34-14-2 'rule', for example);

* I've seen several questions posted on gmatclub about normal distributions, all from the same source, and in almost all of them the math is simply wrong (I'd be curious to know the source of this question);

* the normal distribution is an *infinite* distribution. All of the questions I've seen posted here claim that a *finite* set is 'normally distributed'. That is a mathematical impossibility; a finite set can only be 'approximately normal'. This is the reason why most of these questions don't make any mathematical sense. The explanation below, which I gather you copied from the original source:

Chetangupta

The question tells us that 84% of the students study for at least 114 minutes per night. Remember that the important standard deviation percentages are 34-14-2 ?, and that 114 is one standard deviation above the mean (100 – 2 + 14). Statement 2 tell us that only one out of fifty, or 2%, studies at least 132 minutes; thus 132 is three standard deviations above the mean. Using the two average values we know, we can calculate the difference between each average which equals the standard deviation \(\frac{(132-114)}{3}=6\). Since the mean is one standard deviation below 114, then we know that the mean is 120.

does not make any sense - not only is it consistently using the word 'below' when it means 'above' and uses a faulty equation, but it's also logically wrong. The logical problem in the solution above can be illustrated more simply. If I ask:

Set S contains 50 temperature readings. If the mean of S is 100, what is the standard deviation of S?
1. Exactly one of the values in S is greater than 120.
2. Exactly one of the values in S is more than 2 standard deviations above the mean.

then using both statements, it's certainly possible that the standard deviation of S is 10, but that does not *need* to be true. Maybe the standard deviation is 11, and the value which is greater than 120 is equal to 125, say. The answer here is E.

The same is true in the problem above. Yes, we know that 1 of 50, or 2%, of students studied for longer than 132 minutes. That does not mean the value 132 is exactly two standard deviations above the mean, as the solution above claims. It might be that the value 131.7 is exactly two standard deviations above the mean, and that there are just no values between 131.7 and 132 in the set. So Statement 2 is certainly not sufficient here.
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Thanks Ian, this question is from princeton. I too had marked A but the OA is D. probably some one from princeton can clarify?
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A college surveyed 50 students to determine the average 16 Apr 2015, 09:25
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Hi,

I have gone through the details of SD and able to clear some concepts also. Thanks for sharing the links. However, I am stuck on the second option saying that (2) Only 1 student studied at least 132 minutes per night. 114 is one SD below the mean- I made of it. But, I am unable to interpret how 132 is 2 SD above the mean. How to make sure that 132 lies with 95% since we don't know the mean. Kindly shed some light on it. I am struggling on it for a long time.

Thanking you in advance. :)
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A college surveyed 50 students to determine the average 16 Apr 2015, 09:55
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deya
Hi,

I have gone through the details of SD and able to clear some concepts also. Thanks for sharing the links. However, I am stuck on the second option saying that (2) Only 1 student studied at least 132 minutes per night. 114 is one SD below the mean- I made of it. But, I am unable to interpret how 132 is 2 SD above the mean. How to make sure that 132 lies with 95% since we don't know the mean. Kindly shed some light on it. I am struggling on it for a long time.

Thanking you in advance. :)
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This is a poor quality question on a concept which is NOT tested on the GMAT. Check THIS post for more.

Topic locked.
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A college surveyed 50 students to determine the average 19 Jul 2015, 07:40
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Hello all,

How do we solve this DS Question?

A college surveyed 50 students to determine the average number of minutes that students studied per night. If 84 percent of the students studied for at least 114 minutes per night and the students’ responses had a normal distribution, what was the average number of minutes studied per night?

1.If 10 more students averaging 192 minutes per night were added to the sample, the new average would be 10 percent greater.
2.Only 1 student studied at least 132 minutes per night.

P:S: Do we have any theory/tips and tricks to master GMAT Descrptive Statistics?

Thanks.
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A college surveyed 50 students to determine the average 19 Jul 2015, 07:52
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sinraj
Hello all,

How do we solve this DS Question?

A college surveyed 50 students to determine the average number of minutes that students studied per night. If 84 percent of the students studied for at least 114 minutes per night and the students’ responses had a normal distribution, what was the average number of minutes studied per night?

1.If 10 more students averaging 192 minutes per night were added to the sample, the new average would be 10 percent greater.
2.Only 1 student studied at least 132 minutes per night.

P:S: Do we have any theory/tips and tricks to master GMAT Descrptive Statistics?

Thanks.
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This question has already been discussed here. Please search for your question in the forums before posting it. The topic is merged with another similar locked topic.

a-college-surveyed-50-students-to-determine-the-average-30723.html#p1516312

This Question is Locked Due to Poor Quality
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