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During this year's fundraiser, students who sell at least 75 subscript

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Bunuel
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During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   07 Mar 2019, 02:53
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E

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95% (hard)

Question Stats:

26% (02:05) correct 74% (01:38) wrong based on 27 sessions
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During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given

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C
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sumi747
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   07 Mar 2019, 03:13
16% should be giving 71 subscriptions or greater than that, hence with the given info we cannot determine the percentage of students who would receive the prize.
IMO E

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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   07 Mar 2019, 06:32
Bunuel wrote:
During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given


75=47+14*x
x= 2
IMO C
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   07 Mar 2019, 11:47
Archit3110 wrote:
Bunuel wrote:
During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given


75=47+14*x
x= 2
IMO C


Archit3110
Can you elaborate on what you have done? Cause I used the bell curve to arrive at my answer

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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   08 Mar 2019, 02:33
sumi747 wrote:
Archit3110 wrote:
Bunuel wrote:
During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given


75=47+14*x
x= 2
IMO C


sumi747

I used following formula to determine whether 75 is how much SD above or below the mean
value= mean + SD*x
we need to determine x in this case
so
75=47+14*x
x= 2


hope this helps..
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   11 Mar 2019, 05:33
Archit3110 wrote:
Bunuel wrote:
During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given


75=47+14*x
x= 2
IMO C



Hi

Could you please explain how you arrived at this expression "75=47+14*x" ?

Regards,

Rit
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Archit3110
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   11 Mar 2019, 05:35
menonrit wrote:
Archit3110 wrote:
Bunuel wrote:
During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given


75=47+14*x
x= 2
IMO C



Hi

Could you please explain how you arrived at this expression "75=47+14*x" ?

Regards,

Rit


menonrit
I used following formula to determine whether 75 is how much SD above or below the mean
value= mean + SD*x
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   11 Mar 2019, 05:57
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Bunuel wrote:
During this year's fundraiser, students who sell at least 75 subscriptions will win a prize. The fourth-grade students sold an average of 47 magazine subscriptions per student, and the sales have a standard deviation of 14. If the sales of subscriptions are normally distributed, then approximately what percent of the fourth-grade students will receive a prize?

A. 0.02
B. 0.25
C. 2
D. 25
E. It cannot be determined from the information given


This is what a normal distribution looks like. The sales are normally distributed. This means:
68% sales will lie within 1 SD of the mean.
95% sales will lie within 2 SD of the mean.
Attachment:
normalDistr.jpg
normalDistr.jpg [ 220.26 KiB | Viewed 17386 times ]

75 = Mean + 2*SD = 47 + 2*14
So 5% percent students will lie outside 2 SD - 2.5% on the left (green left region + its left) and 2.5% on the right (green right region + its right).
The ones in the green right region + its left will be those who would have 75 subscriptions or more.
So about 2.5% of the students will receive a prize.

Answer (C)
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink] New post   11 Aug 2023, 07:45
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Re: During this year's fundraiser, students who sell at least 75 subscript [#permalink]
11 Aug 2023, 07:45
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