Question

PLEASE HELP! ITS ALMOST MIDNIGHT AND IM TAKING MY BENCHMARKS

Answer 1

The answer is C. Hope this helps

Answer 2

Answer:

The median number of shoes for the girls is greater than the median number of shoes for the boys.

Step-by-step explanation:

Jeffrey surveys the 13 girls and 13 boys in his class to find out how many pairs of shoes they have in their closets

Girls Data :10, 12, 9 , 15, 25,8, 6,14, 18 , 11 , 9 , 13 , 15

Arrange the data in the ascending order :

6,8,9,9,10,11,12,13,14,15,15,18,25

No. of observations n = 13(odd)

So, Median = $(\frac{n+1}{2})^{\text{th term}}$

                  = $(\frac{13+1}{2})^{\text{th term}}$    

                  = $(\frac{14}{2})^{\text{th term}}$    

                  = $(7)^{\text{th term}}$    

                  = $12$    

Thus the median of the girls data is 12.

Now to find $Q_3$

Consider the set of values right to the median .

13,14,15,15,18,25

Now find the median of this data

No. of observations n = 6(even)

Median = $\frac{(\frac{n}{2}+1)^{\text{th term}}+\frac{n}{2}^{\text{th term}}}{2}$

So, Median = $\frac{(\frac{6}{2}+1)^{\text{th term}}+\frac{6}{2}^{\text{th term}}}{2}$

Median = $\frac{4^{\text{th term}}+3^{\text{rd term}}}{2}$

Median = $\frac{15+15}{2}$

Median = $15$

Mean = $\frac{\text{Sum of all observations }}{\text{number of observations}}$

Mean = $\frac{10+12+ 9 +15+25+8+6+14+18+ 11 +9+ 13+15}{13}$

Mean = $12.69$

So, $Q_3$ for girls is 15 and median is 12 and mean is 12.6.

Boys data : 8,6,4,5,10,15,7,8,12,11,9,5,4

Arrange the data in the ascending order :

4,4,5,5,6,7,8,8,9,10,11,12,15

No. of observations n = 13(odd)

So, Median = $(\frac{n+1}{2})^{\text{th term}}$

                  = $(\frac{13+1}{2})^{\text{th term}}$    

                  = $(\frac{14}{2})^{\text{th term}}$    

                  = $(7)^{\text{th term}}$    

                  = $8$    

Thus the median of the boys data is 8

Now to find $Q_3$

Consider the set of values right to the median .

8,9,10,11,12,15

Now find the median of this data

No. of observations n = 6(even)

Median = $\frac{(\frac{n}{2}+1)^{\text{th term}}+\frac{n}{2}^{\text{th term}}}{2}$

So, Median = $\frac{(\frac{6}{2}+1)^{\text{th term}}+\frac{6}{2}^{\text{th term}}}{2}$

Median = $\frac{4^{\text{th term}}+3^{\text{rd term}}}{2}$

Median = $\frac{11+10}{2}$

Median = $10.5$

Mean = $\frac{\text{Sum of all observations }}{\text{number of observations}}$

Mean = $\frac{4+4+5+5+6+7+8+8+9+10+11+12+15}{13}$

Mean = $8$

So, $Q_3$ for boys is 10.3 and median is 8 and mean is 8

Since we can see that the value of $Q_3$ , mean and median opf girls is greater than boys.

So, Option C is correct.

The median number of shoes for the girls is greater than the median number of shoes for the boys.