nbme 3 block 3 ..q 1 ---------to------q50 - drona99

[ArchivalUser]

Bump....

So wats the answer for Q 49

I think it should be A..

[ArchivalUser]

thx

[ArchivalUser]

************************correction*******************************
posted by Sarim

49.A

correct is "A" and the graph is "positively skewed"

just look carefully on what's the x-axis and y-axis......actually they swapped the axises on the graph.....so place the serum creatinine values on x-axis and the pt on y-axis and now look at the graph.

If u are approaching it as How skewed it look? as the tail is pointing towards left then yes it look like negatively skewed.But according to "Frequency Distribution Graph" the y-axis represent the "FREQUENCY of cases or values" or in other words so and so # of patients(or % of a group or population) that correlate with the "MEASUREMENT VALUES" given on the x-axis.

But on the graph given in the stem the y-axis representing the "Measurement values"(serum creatinin) rather than the Frequency of appearing of those serum creatinin values and x-axis representing the # of Observations (# of patients) rather than Measurement values.

So in order to decide just by looking at the shape of the graph whether the Distribution is normal,+ skewed or - skewed we need to be looking at "Frequency Distribution Graph" and which is not the case in this stem.

Now if calculate this:

Patient #----------- 1---------2-------3------4-------5-------6------7-------8-------9--------10
creatinin values:--0.5------0.6-----0.7----0.8------1-------1------1-------2-------4---------5

MEDIAN:
Since total # of patients is an EVEN #i.e. 10 ,we get middle two values and divide them by 2
patient 5th and 6th got creatinin values of 1 and 1.so it will be 1+1 divided by 2=1 (Median)

Median= 1
Mode = 1 (most frequently appearing value)

Mean = (.5 +.6 +.7 +.8+ 1 +1 + 1 + 2 + 4 + 5) divide by 10 ( total# of patients)
........= 16.6/10
........= 1.6

so
Median=1
Mode =1
Mean =1.6

Mean>Median

Now if we were to place Measurement values(creatinin) in a Frequency Distribution graph then the serum creatinin value of 1 mg/dl appeared in 3 patients and all other values appeared only once ,so there will be one Tall Bar(representing 1 mg/dl appearing 3 times) and remaining seven Smaller Bars of same height cz all seven have different creatinin values but each appeared only once.


-just turn this graph around where Frequencies(# of patients) are on y-axis and creatinin values are on x-axis and when u draw a line at the height of the Bars it will give u a tail pointing towards right (+ skewed)

[ArchivalUser]

Q-20 - D.........Can someone please explain why "D"?

[ArchivalUser]

to add more for Q49....AAA

If u are approaching it as How skewed it look? as the tail is pointing towards left then yes it look like negatively skewed.But according to "Frequency Distribution Graph" the y-axis represent the "FREQUENCY of cases or values" or in other words so and so # of patients(or % of a group or population) that correlate with the "MEASUREMENT VALUES" given on the x-axis.

But on the graph given in the stem the y-axis representing the "Measurement values"(serum creatinin) rather than the Frequency of appearing of those serum creatinin values and x-axis representing the # of Observations (# of patients) rather than Measurement values.

So in order to decide just by looking at the shape of the graph whether the Distribution is normal,+ skewed or - skewed we need to be looking at "Frequency Distribution Graph" and which is not the case in this stem.

Now if calculate this:

Patient #----------- 1---------2-------3------4-------5-------6------7-------8-------9--------10
creatinin values:--0.5------0.6-----0.7----0.8------1-------1------1-------2-------4---------5

MEDIAN:
Since total # of patients is an EVEN #i.e. 10 ,we get middle two values and divide them by 2
patient 5th and 6th got creatinin values of 1 and 1.so it will be 1+1 divided by 2=1 (Median)

Median= 1
Mode = 1 (most frequently appearing value)

Mean = (.5 +.6 +.7 +.8+ 1 +1 + 1 + 2 + 4 + 5) divide by 10 ( total# of patients)
........= 16.6/10
........= 1.6

so
Median=1
Mode =1
Mean =1.6

Mean>Median (as well as Mean>Mode)

Now if we were to place Measurement values(creatinin) in a Frequency Distribution graph then the serum creatinin value of 1 mg/dl appeared in 3 patients and all other values appeared only once ,so there will be one Tall Bar(representing 1 mg/dl appearing 3 times) and remaining seven Smaller Bars of same height cz all seven have different creatinin values but each appeared only once.

**** Quick approach
look at y axis 7 out of 10 values are located in range of 0.5 and 1.0...
only 3 of 10 values are located in range of 2.0 and 5.0...
if you draw a bell shaped curve and imagine that the hump correspond to the most often values.. here most often values are small figures (between 0.5 and 1.0), so the hump of the curve would be in the left, large figures are less often (between 2.0 and 5.0) would be tail of the curve and located in the right...
therefore it is POSITIVE SKEWED CUREVE!!!!!!!

When u swap the axis:

>>>just turn this graph around where Frequencies(# of patients) are on y-axis and creatinin values are on x-axis and when u draw a line at the height of the Bars it will give u a tail pointing towards right (+ skewed)

Frequency (# of patients)
........0........1.......2.......3.......4.......5.....till 10
........___________________________
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|..0.5 |||||||||
|..0.6 |||||||||
|..0.7 |||||||||
|..0.8 |||||||||
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|..1.0 |||||||||||||||||||||||
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|..2.0 ||||||||
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|..4.0 ||||||||
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|..5.0 ||||||||
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posted by Sarim

[ArchivalUser]

pls kindly assist me with nbme 7, 11 and 12 , with answers pleassssssssssssse!. chinex791

[ArchivalUser]

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