Respuesta :
first arrange in ascending order:-
4 5 6 6 7 8 9 9 10 10 12 12 14 15
first quartile = 6 and third quartile = 12
difference = 12-6 = 6
4 5 6 6 7 8 9 9 10 10 12 12 14 15
first quartile = 6 and third quartile = 12
difference = 12-6 = 6
Answer:
The difference of the values of the first and third quartiles of the data set is 6.
Step-by-step explanation:
Data : 10, 9, 6, 12, 4, 6, 7, 8, 15, 14, 12, 9, 10, 5
This data set represents the number of cups of coffee sold in a café between 8 a.m. and 10 a.m. every day for 14 days.
Arrange the data in the ascending order :
4,5,6,6,7,8,9,9,10,10,12,12,14,15
No. of observations n = 14(even)
Median = [tex]\frac{(\frac{n}{2}+1)^{\text{th term}}+\frac{n}{2}^{\text{th term}}}{2}[/tex]
So, Median = [tex]\frac{(\frac{14}{2}+1)^{\text{th term}}+\frac{14}{2}^{\text{th term}}}{2}[/tex]
Median = [tex]\frac{8^{\text{th term}}+7^{\text{th term}}}{2}[/tex]
Median = [tex]\frac{9+9}{2}[/tex]
Median = [tex]\frac{18}{2}[/tex]
Median = 9
Now to find [tex]Q_3[/tex] i.e. Third quartile
Consider the set of values right to the median .
9,10,10,12,12,14,15
Now find the median of this data
No. of observations n = 7(odd)
So, Median = [tex](\frac{n+1}{2})^{\text{th term}}[/tex]
= [tex](\frac{7+1}{2})^{\text{th term}}[/tex]
= [tex](\frac{8}{2})^{\text{th term}}[/tex]
= [tex](4)^{\text{th term}}[/tex]
= [tex]12[/tex]
So, [tex]Q_3=12[/tex]
Now to find [tex]Q_1[/tex] i.e. First quartile
Consider the set of values left to the median .
4,5,6,6,7,8,9
No. of observations n = 7(odd)
So, Median = [tex](\frac{n+1}{2})^{\text{th term}}[/tex]
= [tex](\frac{7+1}{2})^{\text{th term}}[/tex]
= [tex](\frac{8}{2})^{\text{th term}}[/tex]
= [tex](4)^{\text{th term}}[/tex]
= [tex]6[/tex]
So, [tex]Q_1=6[/tex]
Now The difference of the values of the first and third quartiles of the data set : 12-6=6
Hence The difference of the values of the first and third quartiles of the data set is 6
