given the median , find an unknown
$begingroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
$endgroup$
add a comment |
$begingroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
$endgroup$
add a comment |
$begingroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
$endgroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
edited Mar 31 '16 at 19:20
V. Vancak
11.4k3926
11.4k3926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
asked Mar 31 '16 at 7:52
user307640user307640
8691719
asked Mar 31 '16 at 7:52
user307640user307640
8691719
8691719
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
add a comment |
$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
add a comment |
$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
answered Mar 31 '16 at 8:38
NikolasNikolas
628
answered Mar 31 '16 at 8:38
NikolasNikolas
628
answered Mar 31 '16 at 8:38
NikolasNikolas
628
628
add a comment |
add a comment |
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