Question

Suppose that the following frequency table contains data on the number of absences in a school year due to illness or injury for 80 randomly selected high school students. Use the table to determine the median number of absences for these 80 students. Provide your answer with precision to one decimal place.

Number of absences	Frequency
0	20
1	9
2	11
3	8
4	5
5	10
6	4
7	3
8	3
9	1
10	3
11	3

Median = _____ Absences

Scores on a university papers are Normally distributed, with a mean of 7878 and a standard deviation of 88 . The professor teaching the class declares that a score of 7070 or higher is required for a grade of at least C. Using the 68–95–99.768–95–99.7 rule, what percent of students failed to earn a grade of at least C?

A. 32%

B. 16%

C. 5%

Answer 1

The table given below ,

Number of absences(x)	Frequency	Cumulative frequency (cf)
0	20	20
1	9	20+9=29
2	11	29+11=40
3	8	40+8=48
4	5	48+5=53
5	10	53+10=63
6	4	63+4=67
7	3	67+3=70
8	3	70+3=73
9	1	73+1=74
10	3	74+3=77
11	3	=77+3=80
	N=80

Now , the median is the observation corresponding to the cumulative frequency and that cumulative frequency is just greater than or equal (N/2)

Here ,

Therefore , the observation corresponding cf=40 is the median.

Therefore , median=2 Absences.