Question

question
a principal of a high school wanted to study the attendance of the student body. the principal looked at a random sample of 10 students and recorded their absences. the end of year absences for the students are reproduced in the table below. calculate the mean, median, and mode using a ti - 83, ti - 83 plus, or ti - 84 graphing calculator (round your answers to one decimal place).
school days
missed
3
1
8
7
3
1
5
3
9
3
provide your answer below:
mean = \\(\square\\), median = \\(\square\\), mode = \\(\square\\)

Explanation:

Step1: Calculate the Mean

First, sum all the values: \(3 + 1 + 8 + 7 + 3 + 1 + 5 + 3 + 9 + 3\). Let's compute that: \(3+1 = 4\), \(4 + 8 = 12\), \(12 + 7 = 19\), \(19 + 3 = 22\), \(22 + 1 = 23\), \(23 + 5 = 28\), \(28 + 3 = 31\), \(31 + 9 = 40\), \(40 + 3 = 43\). Then divide by the number of data points (10). So the mean is \(\frac{43}{10}=4.3\).

Step2: Calculate the Median

First, order the data: \(1, 1, 3, 3, 3, 3, 5, 7, 8, 9\). There are 10 data points, so the median is the average of the 5th and 6th values. The 5th value is 3 and the 6th value is 3. So the median is \(\frac{3 + 3}{2}=3.0\)? Wait, no, wait, let's re - order: Wait, the data set is [1,1,3,3,3,3,5,7,8,9]? Wait, no, wait the original data: 3,1,8,7,3,1,5,3,9,3. Let's sort them: 1,1,3,3,3,3,5,7,8,9. Wait, 10 numbers. The 5th and 6th terms: the 5th is 3, the 6th is 3. So median is \(\frac{3 + 3}{2}=3.0\)? Wait, no, wait, wait, 1,1,3,3,3,3,5,7,8,9. Wait, 1 (1st), 1 (2nd), 3 (3rd), 3 (4th), 3 (5th), 3 (6th), 5 (7th), 7 (8th), 8 (9th), 9 (10th). So the median is the average of the 5th and 6th, which are both 3. So median is 3.0? Wait, but let's check the sum again. Wait, 1+1 = 2, +3 = 5, +3 = 8, +3 = 11, +3 = 14, +5 = 19, +7 = 26, +8 = 34, +9 = 43. Yes, sum is 43. Mean is 4.3. Now mode: the number that appears most often. 3 appears 4 times, which is more than any other number (1 appears 2 times, others once). So mode is 3.

Wait, wait, I think I made a mistake in the median calculation. Wait, the sorted data: 1,1,3,3,3,3,5,7,8,9. The 5th term is 3, the 6th term is 3. So median is (3 + 3)/2 = 3.0? But wait, let's check the count again. Wait, the data points: 1 (2 times), 3 (4 times), 5 (1), 7 (1), 8 (1), 9 (1). So sorted: [1,1,3,3,3,3,5,7,8,9]. So positions 1 - 10. The median for even number of data points is the average of the n/2 and (n/2 + 1)th terms. n = 10, so n/2 = 5, n/2+1 = 6. The 5th term is 3, 6th term is 3. So median is (3 + 3)/2 = 3.0? But wait, when I calculated the mean, sum is 43, 43/10 = 4.3. Mode is 3 (appears 4 times).

Wait, but let's re - check the sum: 1+1 = 2; 2+3 = 5; 5+3 = 8; 8+3 = 11; 11+3 = 14; 14+5 = 19; 19+7 = 26; 26+8 = 34; 34+9 = 43. Yes, sum is 43. So mean is 4.3. Median: sorted data, 5th and 6th terms are 3 and 3, so median is 3.0? Wait, no, wait, 1,1,3,3,3,3,5,7,8,9. The 5th term is the 5th number: index 4 (0 - based) or 5 (1 - based). Wait, 1 - based: 1 (1), 1 (2), 3 (3), 3 (4), 3 (5), 3 (6), 5 (7), 7 (8), 8 (9), 9 (10). So 5th is 3, 6th is 3. So median is (3 + 3)/2 = 3.0. Mode is 3.

Answer:

Mean = \(4.3\), Median = \(3.0\), Mode = \(3\)