Question

A school is tracking its freshmen attendance for the first marking period. Shown below is a table summarizing the findings for the 284 members of the freshmen class.
(a) Find the mean and median number of days absent Round your mean to the nearest tenth

(b) What proportion of the population that has an absenteeism greater than 4 days?​

Answer 1

Answer:

The median number of days absent is zero (0)

The mean number of days absent is 1.0 day

(b) The proportion of the population that has absenteeism greater than 4 days is 6.34 %

Step-by-step explanation:

total number of students, n = 284

The total number of students is even, the median number of days absent will in (n/2).

n/2 = 284/2 = 142

The cumulative frequency that falls in 148 students = 0 day

The median number of days absent is zero (0)

For mean:

Let the days absent = x  

let the number of students = f

$mean (\bar x) = \frac{\sum fx}{\sum f}\\\\\bar x = \frac{(158\times 0)+ (64\times 1)+(18\times 2)+(22\times 3)+(4\times 4)+(5\times 7)+(6\times 8)+(9\times 2)+(13\times 1)}{284} \\\\\bar x = \frac{296}{284} \\\\\bar x = 1.0 \ day$

(b) the number of students with absenteeism greater than 4 dyas;

= 7 + 8 + 2 + 1

= 18

The proportion of these students;

$= \frac{18}{284} \\\\= 0.0634 \\\\= 6.34 \ \%$