NBME 4_1 17

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NBME 4_1 17

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DrVirgo
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06/02/06 - 02:02 PM

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mjl1717
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06/02/06 - 02:50 PM

45%

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sridevibandaru24
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06/03/06 - 12:20 PM

45%

nadiabarati

06/03/06 - 07:29 PM

45%

DrVirgo
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06/03/06 - 07:38 PM

agree 10 + 35. smiling face

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Cedrick
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06/05/06 - 06:30 PM

Diego Casali
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06/06/06 - 11:17 AM

robin082006
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06/06/06 - 11:26 AM

Agree

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asankineni
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06/21/06 - 11:07 AM

Does step 1 give such a simple question. Its more like GRE question

docarchana
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09/27/06 - 07:39 PM

#10

it is 10%
10 out of 100 (25+30+10+35) have two episodes..

klimt
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10/01/06 - 09:52 AM

#11

yes but the 35 with 3 episodes also have at least 2, right? wink

so it's 10+35=45%
E

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IceAge
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10/01/06 - 08:47 PM

#12

hey !!!!!

this one is like "how many months have 28 d in them?" ans: "all 12!!!"

kishore80
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10/06/06 - 07:52 PM

#13

thread extension:

what is the average number of UTIs in these children?

a) between 0 and1

b) between 1 and 2

c) btw 2 and 3

d) cant be determined

vamsi2004
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10/07/06 - 08:44 PM

#14

i guess its between 0 and 1..

robin082006
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10/07/06 - 09:38 PM

#15

= (25*0 + 30*1 + 10*2 + 35*3)= 155/100

answer is B

kishore80 wrote:
thread extension:

what is the average number of UTIs in these children?

a) between 0 and1

b) between 1 and 2

c) btw 2 and 3

d) cant be determined

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vamsi2004
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10/07/06 - 11:19 PM

#16

i think its btw 0 and 1 bcos some of the numbers in those two pillars (25+30) is greater than that btw 1 and 2 (30+10)

young_doc
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10/08/06 - 01:31 PM

#17

robin082006 wrote:
= (25*0 + 30*1 + 10*2 + 35*3)= 155/100

answer is B

I'm not sure we can multiply the number of children by the number of episodes???

It's like if i say odds of getting a Jack(0) is 25%, odds of getting a Queen(1) is 30%, odds of getting a King(2) is 10%, and odds of getting an Ace(3) is 35%.

Well then we're more likely to get a mix of Jacks & Queens (i.e. 0 to 1), instead of Quens & Kings (1 to 2), or Kings & Aces (2 to 3).

By your reasoning Robin you would be doing [25xJack + 30xQueen.....]

This is of course just an analogy, but i think you understand my confusion as to why you're multiplying the number children by the number of episodes???

Thanks in advance Robin for an explanation of your calculations.

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study_ing
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10/09/06 - 05:46 AM

#18

i agree with B and Robin's calculations. an average wud simply be the total number of episodes dived by the total number of children.

hence to calculate the total number of episodes, u need to multiply eg

2 episodes /child *number of children,

add them up to get a total.

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Luckyall
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10/09/06 - 10:06 AM

#19

i did it in a different way.... the Qs clearly asks wht %of kids had MINIMUM 2 episodesUTI ???
From this graph, just observing it, i see that 2 episodes UTI in 10 number of kids... great! Now i just have to figure it out of % of sample these 10 number of kids represent ! 40 number kids------->100%
10 number of kids---->25%
The answer is 10kids had MINIMUM 2 episodes UTI-----> equivalent od 25% of sample had minimum 2 epis UTI!

Addicudo
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10/09/06 - 06:41 PM

#20

Luckyall wrote:
i did it in a different way.... the Qs clearly asks wht %of kids had MINIMUM 2 episodesUTI ???
From this graph, just observing it, i see that 2 episodes UTI in 10 number of kids... great! Now i just have to figure it out of % of sample these 10 number of kids represent ! 40 number kids------->100%
10 number of kids---->25%
The answer is 10kids had MINIMUM 2 episodes UTI-----> equivalent od 25% of sample had minimum 2 epis UTI!

Well here .. the Total number of children is not 40 .. ITS 100 .. !! And they are asking the nuber of children who had a minimum of TWO episodes .. therefore children with three will also be considered .. thus .. making it ..

10 (Number of children who had TWO episodes) +
35 (Number of children who had THREE episodes) DIVIDED by the total no of children .. = 45/100 .. = 45% .. !!

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Luckyall
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10/13/06 - 02:18 PM

#21

Absolutelly correct ! Makes perfect sense! Thanks nod

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