Question

here are the scores of 13 students on a science test.
65, 66, 67, 72, 72, 75, 76, 77, 78, 81, 81, 91, 92
notice that the scores are ordered from least to greatest.
give the five - number summary and the interquartile range for the data set.
five - number summary
minimum:
lower quartile:
median:
upper quartile:
maximum:
interquartile range:

Explanation:

Step1: Find Minimum and Maximum

The data set is ordered: 65, 66, 67, 72, 72, 75, 76, 77, 78, 81, 81, 91, 92.
Minimum is the first value: \( 65 \).
Maximum is the last value: \( 92 \).

Step2: Find Median (Q2)

There are \( n = 13 \) values (odd). Median is the \( \frac{n + 1}{2} = 7 \)-th term.
7th term: \( 76 \).

Step3: Find Lower Quartile (Q1)

Lower half: first 6 values (since median is 7th, lower half is 1–6th: 65, 66, 67, 72, 72, 75).
Q1 is the median of lower half: \( \frac{67 + 72}{2} = 69.5 \)? Wait, no—wait, \( n = 6 \) (even), so median is average of 3rd and 4th. Wait, lower half: positions 1–6 (values: 65, 66, 67, 72, 72, 75). 3rd term: 67, 4th term: 72. Wait, no—wait, original data: 13 terms. Lower half is first 6 terms (before median). Wait, no: when \( n \) is odd, lower half is \( \frac{n - 1}{2} = 6 \) terms (positions 1–6), upper half is positions 8–13 (6 terms). Wait, median is position 7. So lower half: 65, 66, 67, 72, 72, 75 (6 terms). Q1 is median of these 6: average of 3rd and 4th. 3rd: 67, 4th: 72. So \( \frac{67 + 72}{2} = 69.5 \)? Wait, no, wait—wait, maybe I made a mistake. Wait, the data is 13 terms: indices 1 to 13. Median at 7 (76). Lower half: indices 1–6 (values 65,66,67,72,72,75). Upper half: indices 8–13 (values 77,78,81,81,91,92). Wait, no—index 8 is 77, index 9:78, 10:81, 11:81, 12:91, 13:92. Wait, upper half is 6 terms (indices 8–13). So Q1: median of lower half (indices 1–6). 6 terms: median is average of 3rd and 4th. 3rd term: 67, 4th term: 72. So Q1 = (67 + 72)/2 = 69.5? Wait, but let's check again. Wait, maybe the lower half is considered as first 6 terms, but maybe the formula is different. Wait, another approach: for a data set with \( n \) values, Q1 is the value at \( \frac{n + 1}{4} \)-th term (for odd \( n \))? Wait, no, the standard method for quartiles:

For \( n = 13 \):

- Minimum: 65
- Q1: value at \( \frac{n + 1}{4} = 3.5 \)-th term. So average of 3rd and 4th terms. 3rd term: 67, 4th term: 72. So \( \frac{67 + 72}{2} = 69.5 \). Wait, but maybe the problem expects using the median of the lower half (6 terms) as Q1. Wait, lower half has 6 terms (positions 1–6), so median of lower half is (67 + 72)/2 = 69.5.

- Median (Q2): 7th term, 76.

- Q3: value at \( \frac{3(n + 1)}{4} = 10.5 \)-th term. Average of 10th and 11th terms. 10th term: 81, 11th term: 81. So \( \frac{81 + 81}{2} = 81 \). Alternatively, upper half: positions 8–13 (values 77,78,81,81,91,92). Median of upper half: average of 3rd and 4th (81 and 81), so 81.

- Maximum: 92.

- Interquartile range (IQR) = Q3 - Q1 = 81 - 69.5 = 11.5? Wait, no—wait, maybe I messed up Q1. Wait, let's re-express the data:

Data: [65, 66, 67, 72, 72, 75, 76, 77, 78, 81, 81, 91, 92]

n = 13.

Median (Q2) is the 7th term: 76.

Lower half (before Q2): [65, 66, 67, 72, 72, 75] (6 terms). Q1 is the median of this: (67 + 72)/2 = 69.5.

Upper half (after Q2): [77, 78, 81, 81, 91, 92] (6 terms). Q3 is the median of this: (81 + 81)/2 = 81.

So:

Minimum: 65

Lower quartile (Q1): 69.5? Wait, but maybe the problem uses a different method (e.g., inclusive vs exclusive). Wait, maybe I made a mistake. Wait, let's check with another approach. For n=13, the positions for quartiles:

Q1: position \( \frac{n + 1}{4} = 3.5 \) → average of 3rd and 4th term. 3rd term: 67, 4th term: 72 → (67 + 72)/2 = 69.5.

Q3: position \( \frac{3(n + 1)}{4} = 10.5 \) → average of 10th and 11th term. 10th term: 81, 11th term: 81 → 81.

Median: 7th term: 76.

Maximum: 92.

IQR: 81 - 69.5 = 11.5.

Wait, but let's verify with the data:

Wait, maybe the lower quartile is 67? No, because the lower half has 6 terms, so median of 6 terms is average of 3rd and 4th. 3rd term is 67, 4th is 72. So 69.5.

Alternatively, maybe the problem considers the lower half as first 6 terms (indices 1–6) and takes the median as the 3rd term? No, that's for odd numbers. Wait, 6 terms: even, so median is average of 3rd and 4th.

So:

Minimum: 65

Lower quartile (Q1): 69.5

Median: 76

Upper quartile (Q3): 81

Maximum: 92

Interquartile range: 81 - 69.5 = 11.5

Wait, but maybe I made a mistake. Let's check again.

Wait, the data is: 65, 66, 67, 72, 72, 75, 76, 77, 78, 81, 81, 91, 92.

Number of terms: 13.

Median (middle term) is the 7th term: 76. Correct.

Lower half: terms 1–6: 65, 66, 67, 72, 72, 75.

To find Q1, we take the median of these 6 terms. Since there are 6 terms (even), the median is the average of the 3rd and 4th terms.

3rd term: 67

4th term: 72

So Q1 = (67 + 72)/2 = 69.5.

Upper half: terms 8–13: 77, 78, 81, 81, 91, 92.

Median of these 6 terms: average of 3rd and 4th terms.

3rd term: 81

4th term: 81

So Q3 = (81 + 81)/2 = 81.

Minimum: 65

Maximum: 92

IQR: Q3 - Q1 = 81 - 69.5 = 11.5.

Answer:

Five - number summary:
Minimum: \( 65 \)
Lower quartile: \( 69.5 \)
Median: \( 76 \)
Upper quartile: \( 81 \)
Maximum: \( 92 \)
Interquartile range: \( 11.5 \)