Question

the data set represents the total number of tuna players in each of 11 different school teams.
0, 1, 3, 3, 4, 4, 4, 5, 6, 6, 8
what is the lower quartile of the data?
a. 1
b. 3
c. 4
d. 6

Explanation:

Step1: Order the data

First, we order the given data set: \(0, 1, 3, 3, 4, 4, 4, 5, 6, 6, 8\) (Wait, wait, the original data seems to have a typo? Wait, the user wrote "0, 1, 3, 3, 4, 4, 4, 5, 6, 6, 8" but the count: let's check the number of data points. Wait, the problem says 11 different school teams? Wait, maybe the data is \(0, 1, 3, 3, 4, 4, 4, 5, 6, 6, 8\) – no, that's 11 numbers? Wait, 0,1,3,3,4,4,4,5,6,6,8 – let's count: 0 (1),1(2),3(3),3(4),4(5),4(6),4(7),5(8),6(9),6(10),8(11). Yes, 11 data points.

Step2: Find the median (which is the lower quartile's position? Wait, lower quartile (Q1) is the median of the lower half. First, find the median (middle value) of the entire data set. For n=11, the median is at position \(\frac{n+1}{2} = 6\)th term. Let's list the ordered data:

Index: 1:0, 2:1, 3:3, 4:3, 5:4, 6:4, 7:4, 8:5, 9:6, 10:6, 11:8

Median is the 6th term: 4.

Now, the lower half is the data points before the median (positions 1 - 5, since median is position 6). So lower half: 0, 1, 3, 3, 4.

Now, find the median of this lower half. For n=5 (odd), the median is at position \(\frac{5+1}{2} = 3\)rd term.

Lower half data: 0 (1),1(2),3(3),3(4),4(5). The 3rd term is 3. Wait, but wait, maybe I made a mistake. Wait, the original data: let's recheck the data. The user wrote: "0, 1, 3, 3, 4, 4, 4, 5, 6, 6, 8" – yes, 11 numbers.

Wait, but maybe the data is 0,1,3,3,4,4,4,5,6,6,8 (11 numbers). So median is the 6th term: 4. Then lower half is first 5 numbers: 0,1,3,3,4. Median of lower half (Q1) is the 3rd term, which is 3. Wait, but the options are 1,3,4,6. So 3 is an option (option B? Wait, the options are: A.1, B.3, C.4, D.6. So Q1 is 3.

Wait, let's confirm:

Ordered data: [0, 1, 3, 3, 4, 4, 4, 5, 6, 6, 8]

n = 11.

Position of Q1: \(\frac{n+1}{4} = \frac{12}{4} = 3\)rd term? Wait, another method: for n=11, the lower quartile is at the (n+1)/4 th term? Wait, no, the formula for quartiles: for a set with n elements, the lower quartile (Q1) is the value at the \(\lceil \frac{n}{4}
ceil\) th term? Wait, maybe different methods. Let's use the method of dividing the data into four equal parts.

n=11, so each part has approximately 11/4 = 2.75 elements. So Q1 is at the 3rd term (since 2.75 rounded up is 3). Let's check:

First quartile (Q1) is the value where 25% of the data is below it. So for 11 data points, 25% of 11 is 2.75, so the 3rd term. Let's list the data with indices:

1:0, 2:1, 3:3, 4:3, 5:4, 6:4, 7:4, 8:5, 9:6, 10:6, 11:8

So the 3rd term is 3. So Q1 is 3.

Answer:

B. 3