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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?​

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 Cl class=

Answer :

ONYEBUCHINNAJIGO

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

[tex]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[/tex]

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

= 7 + 8 + 2 + 1

= 18

The proportion of these students;

[tex]= \frac{18}{284} \\\\= 0.0634 \\\\= 6.34 \ \%[/tex]

Using statistical concepts, it is found that:

  • a) The mean number of days absent is of 1, while the median is of 0.
  • b) 0.0634 = 6.34% of the population has an absenteeism greater than 4 days.

Mean and Median:

  • The mean is given the sum of the multiplication of each value and it's number of observations, divided by the number of observations.
  • The median is the number of days absent for which the sum of the number of students will be half of the total number, that is, half of 284, which is 142.

Question a:

Then, the mean is:

[tex]M = \frac{0(158) + 1(64) + 2(18) + 3(22) + 4(4) + 5(7) + 6(8) + 9(2) + 13(1)}{284} = 1[/tex]

158 > 142, hence, the median is 0.

Question b:

  • In the sample, there are 284 students.
  • Of those, 7 + 8 + 2 + 1 = 18 were absent four more than 4 days, hence:

[tex]p = \frac{18}{284} = 0.0634[/tex]

0.0634 = 6.34% of the population has an absenteeism greater than 4 days.

To learn more about mean and median, you can take a look at https://brainly.com/question/24732674

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