Question

the heights (in cm) of ten players on a basketball team are shown below. 206, 203, 185, 201, 213, 196, 203, 202, 198, 191 what is the value of the third quartile for this data set? \bigcirc 196 \bigcirc 203 \bigcirc 204.5 \bigcirc 201

Explanation:

Step1: Order the data set

First, we need to order the given data set from smallest to largest. The data points are: 185, 191, 196, 198, 201, 202, 203, 203, 206, 213.

Step2: Find the position of the third quartile (Q3)

The formula for the position of the \( n \)-th quartile is \( i=\frac{3(n + 1)}{4} \) for a data set with \( n \) observations. Here, \( n = 10 \). So, \( i=\frac{3(10 + 1)}{4}=\frac{33}{4}=8.25 \).

Step3: Calculate Q3

Since \( i = 8.25 \), we take the 8th value and 0.25 times the difference between the 9th and 8th values. The 8th value is 203, the 9th value is 206. So, \( Q3=203+0.25\times(206 - 203)=203 + 0.25\times3=203 + 0.75 = 203.75 \)? Wait, no, maybe I used the wrong formula. Another common formula for quartiles when \( n \) is the number of data points: for Q3, the position is \( i=\frac{3n}{4} \) when \( n \) is a multiple of 4, otherwise, we round up. Wait, let's list the data in order: 185, 191, 196, 198, 201, 202, 203, 203, 206, 213. \( n = 10 \). The position of Q3 is \( \frac{3\times10}{4}=7.5 \). Wait, now I'm confused. Wait, the correct method for quartiles (using the method where we split the data into lower half and upper half):

The data set has 10 values, so the median (Q2) is the average of the 5th and 6th values: \( \frac{201 + 202}{2}=201.5 \). Then the upper half (values above the median) are: 202, 203, 203, 206, 213? Wait no, wait the ordered data is: 185, 191, 196, 198, 201, 202, 203, 203, 206, 213. The median is between the 5th (201) and 6th (202) values, so the lower half is the first 5 values: 185, 191, 196, 198, 201. The upper half is the last 5 values: 202, 203, 203, 206, 213. Now, the third quartile is the median of the upper half. The upper half has 5 values, so the median of the upper half is the 3rd value of the upper half. The upper half ordered: 202, 203, 203, 206, 213. The 3rd value is 203? Wait, no, 5 values: positions 1:202, 2:203, 3:203, 4:206, 5:213. The median of the upper half (Q3) is the 3rd value, which is 203? But that's not one of the options. Wait, maybe I made a mistake in ordering. Wait the original data: 206, 203, 185, 201, 213, 196, 203, 202, 198, 191. Let's re - order them correctly: 185, 191, 196, 198, 201, 202, 203, 203, 206, 213. Yes, that's correct. Now, using the formula for quartiles: \( Q_k \) position is \( i=\frac{k(n + 1)}{4} \), \( k = 3 \), \( n = 10 \). So \( i=\frac{3\times11}{4}=8.25 \). So the 8th value is 203, the 9th is 206. So \( Q3=203+0.25\times(206 - 203)=203 + 0.75 = 203.75 \). But the options are 196, 203, 204.5, 201. Wait, maybe the question uses a different method. Wait, let's check the options. The options are 196, 203, 204.5, 201. Wait, maybe I ordered the data wrong. Wait the original data: 206, 203, 185, 201, 213, 196, 203, 202, 198, 191. Let's re - order again: 185, 191, 196, 198, 201, 202, 203, 203, 206, 213. Now, let's use the formula \( Q3=\frac{3(n + 1)}{4} \)-th term. \( n = 10 \), so \( \frac{3\times11}{4}=8.25 \). So the 8th term is 203, the 9th is 206. So 203+0.25*(206 - 203)=203.75. But that's not an option. Wait, maybe the question uses the formula \( Q3=\frac{3n}{4} \)-th term, rounded up. \( 3\times10/4 = 7.5 \), so we take the 8th term? The 8th term is 203. But 203 is an option. Wait, maybe the data was misread. Wait the original data: 206, 203, 185, 201, 213, 196, 203, 202, 198, 191. Wait, maybe I missed a number? Wait the problem says "ten players", so 10 numbers. Wait 185, 191, 196, 198, 201, 202, 203, 203, 206, 213. That's 10 numbers. Wait, maybe the question has a typo, or I made a mistake. Wait, let's check the options again. The options are 196, 203, 204.5, 201. Wait, maybe the data is 206, 203, 185, 201, 213, 196, 203, 202, 198, 191, and another number? No, the problem says ten players. Wait, maybe I used the wrong method. Let's try the method where we calculate Q3 as the median of the upper half. The upper half: the data above the median. The median is between 201 and 202, so the upper half is 202, 203, 203, 206, 213? No, the median is (201 + 202)/2 = 201.5, so the upper half is the values greater than or equal to 201.5? So 202, 203, 203, 206, 213. The median of this upper half is the 3rd value, which is 203. So Q3 is 203. So the answer is 203.

Answer:

203