The following table shows a frequency distribution of the scores obtained in a test.
Punctuation (3, 4] (4, 5] (5, 6] (6, 7] (7, 8] (8, 9] (9, 10]
Number of participants 2 4 10 20 40 35 9
(a) The highest score reached by the bottom 20% of the participants and the score lowest obtained by the top 25% of the participants
(b) The quartiles of the distribution
For section a) to be continuous variables that can have any value within a range what I do is calculate the average for that 20%, that is to say who is in position 24. I calculate the relative frequency (that for this I need to know the absolute frequency) and multiplied it by the midpoint of the interval. I am applying according to the definition and the formula, but I am stuck.
It is a note that is between (6,7] and the average I get 0.1, so the note would be 6.1.
For b) of the quartiles I get that 1 is in (6,7], 2 is in (7,8] and 3 is in (8,9), but I do not know how to get the exact note
I've done it that way, but I do not know if it's okay since it's the first time I've done something similar, can someone verify it?
a) I have the table
Class | partici | Ni
3-4 | 2 | 2
4-5 | 4 | 6
5-6 | 10 | 16
6-7 | 20 | 36
7-8 | 40 | 76
8-9 | 35 | 111
9-10 | 9 | 120
For The highest score reached by the bottom 20% of the participants means the 20th percentile so
P20 = 6+ ((0,20*120)-16)*1 /20 = 6,4 score
The lowest score obtained by the top 25% of the participants means the 75th percentile, so it is calculated in the same way
P75 = 8 + ((0,75*120)-76)*1 /35 = 8,4 score
b)
For the quartiles, it is the same as calculating the percentiles for 25% 50% and 75%. Q2 is equal to the median
Q1 = 6 + ((0,25*120)-16)*1 /20 = 6,7 score
Q2 = 7 + ((0,5*120)-36)*1 /40 = 7,6 score
Q3 = 8 + ((0,75*120)-76)*1 /35 = 8,4 score =P75 (From the section a-))
(6, 7], not(6,7]. In JaX, you can use$(6,\; 7]$to get $(6,; 7].$ – BruceET Sep 14 '18 at 17:34