Maths:median of ungrouped data

1.
mean = (1/8)[2 + (-1) + 4 + 6 + 9 + (-4) + 0 + y]
2.5 = (1/8)[2 + (-1) + 4 + 6 + 9 + (-4) + 0 + y]
2.5 * 8 = 2 - 1 + 4 + 6 + 9 - 4 + y
20 = 16 + y
y = 4
now i'm going to line the no.s from the smaller one to the lager one
-4 , -1 , 0 , 2 , 4 , 4 , 6 , 9
we got that the median is (2 + 4)(1/2) = 3
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2.
sorry , i am not really sure that i understand what is the answer of question 2 ask for . so i'm going to give you the mean , median , Q1 and Q3 , the range , the range of Q1 and Q3 .
the mean is = (0 * 7 + 1 * 9 + 2 * 5 + 3 * 3 + 4 * 1)(7 + 9 + 5 + 3 + 1)
= 32 / 25
= 1.28
the median is = 1
Q1 = 0
Q3 = 3
the range = 4 - 0 = 4
the range of Q1 and Q3 = 3 - 0 = 3
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p.s. if you want to know any other answer of question 2 you may ask me . i hope the answer can hope you .

1.) The total value of these six numbers is 2.5*8=20
2+(-1)+4+6+9+(-4)+0+y=20
y=4

2.) frequency is also known as times (次數)
Therefore<p>

there are 7 days with 0 student absent
there are 9 days with 1 student absent
there are 5 days with 2 students absent
there are 3 days with 3 students absent
there are 1 day with 4 students absent
And there are 9*1+5*2+3*3+1*4=32 abentees in a class in 25
school days
(PLZ WRITE THE QUESTION IN DETAIL !!)