I need help to solve this question

A

AhmedFathy

Banned
Joined
Nov 21, 2014
Messages
4
The following table represents the absent days of a sample of students in the statistics course
Absent days01234
Number of students1058Y3
Determine the value of Y if the median of the absent days is 1.5 day.

Thanks in advance
 
AhmedFathy said:
The following table represents the absent days of a sample of students in the statistics course
Absent days01234
Number of students1058Y3
Determine the value of Y if the median of the absent days is 1.5 day.

Thanks in advance
Click to expand...

In my opinion, this problem has not been stated correctly.

Median of "absent days" is 2 - not 1.5 - and it does not depend on the value of "Y", which belongs to another data set.

Also the "number of students" set is confusing, to some extent. It looks like the students who were "absent for 2 days" were not included in the set "absent for 1 day".

I thought the author wanted students to calculate the value of "Y" using 1.5 = (10*0 + 1*5 + 2*8 + 3*Y + 4*3)/(10+5+8+Y+3) - however that produces fractional value of Y.

So.....
 
Last edited by a moderator:
I have to agree with Subohotosh Khan here. This is just very simple arithmetic if you know the definition of "median". Look it up in your text book.
Click to expand...

Added after Subhotosh Khan changed his response: Now I have to disagree! First, that last computation, 1.5 = (10*0 + 1*5 + 2*8 + 3*Y + 4*3)/(10+5+8+Y+3), was for the mean, not median. The "median" is the "middle number" in the sense that half the data lies below it, half above. (That's the "definition" you needed to know, AhmedFathy.) With the data given here, that means we must have 10+ 5, the number of cases with "absent days" less than 1.5, equal to 8+ y+ 3, the number of cases with "absent days" larger than 1.5. That is, solve 10+ 5= 8+ y+ 3 for y.

(Strictly speaking, since "absent days" is an integer, with that value of y, the "median" could be any number between 1 and 2. By convention, if the median is between two of the given values, it is half way between.)
 
Last edited:
HallsofIvy said:
Added after Subhotosh Khan changed his response: Now I have to disagree! First, that last computation, 1.5 = (10*0 + 1*5 + 2*8 + 3*Y + 4*3)/(10+5+8+Y+3), was for the mean, not median. The "median" is the "middle number" in the sense that half the data lies below it, half above. (That's the "definition" you needed to know, AhmedFathy.) With the data given here, that means we must have 10+ 5, the number of cases with "absent days" less than 1.5, equal to 8+ y+ 3, the number of cases with "absent days" larger than 1.5. That is, solve 10+ 5= 8+ y+ 3 for y.

(Strictly speaking, since "absent days" is an integer, with that value of y, the "median" could be any number between 1 and 2. By convention, if the median is between two of the given values, it is half way between.)
Click to expand...

Subhotosh Khan said:
In my opinion, this problem has not been stated correctly.

Median of "absent days" is 2 - not 1.5 - and it does not depend on the value of "Y", which belongs to another data set.

Also the "number of students" set is confusing, to some extent. It looks like the students who were "absent for 2 days" were not included in the set "absent for 1 day".

I thought the author wanted students to calculate the value of "Y" using 1.5 = (10*0 + 1*5 + 2*8 + 3*Y + 4*3)/(10+5+8+Y+3) - however that produces fractional value of Y.

So.....
Click to expand...

.
 
Last edited:
You must log in or register to reply here.
Top