Question

find the 5 number summary for the data shown
31 33
34 49
53 54
55 56
64 71
72 83
89 92
96
5 number summary:
□, □, □, □, □
use the locator/percentile method described in your book, not your calculator.

Explanation:

Step1: List and sort the data

First, we list all the data points: 31, 33, 34, 49, 53, 54, 55, 56, 64, 71, 72, 83, 89, 92, 96. Then we sort them (they are already in order here). The number of data points \( n = 15 \).

Step2: Find the minimum

The minimum value is the smallest number in the dataset. Looking at the sorted data, the minimum is 31.

Step3: Find the first quartile (Q1)

The locator for Q1 is \( L_{25}=\frac{25}{100}(n + 1)=\frac{25}{100}(15 + 1)=\frac{25}{100}\times16 = 4 \). Since the locator is an integer, Q1 is the average of the 4th and 5th values? Wait, no, wait. Wait, the formula for the locator/percentile method: for a percentile \( P \), the locator \( L_P=\frac{P}{100}(n + 1) \). Wait, actually, some sources use \( L_P=\frac{P}{100}(n) \), but the problem says to use the method in the book. Let's check the standard locator method. Let's confirm the number of data points: let's count the numbers. The table has: first row 2, second 2, third 2, fourth 2, fifth 2, sixth 2, seventh 2, eighth 1. So total \( 2+2+2+2+2+2+2+1 = 15 \) data points. So \( n = 15 \).

For the minimum (0th percentile), \( L_0=\frac{0}{100}(15 + 1)=0 \), but the minimum is the first value, 31.

For the median (50th percentile), \( L_{50}=\frac{50}{100}(15 + 1)=\frac{1}{2}\times16 = 8 \). So the median is the 8th value. Let's list the positions: 1:31, 2:33, 3:34, 4:49, 5:53, 6:54, 7:55, 8:56. So median (Q2) is 56.

For Q1 (25th percentile), \( L_{25}=\frac{25}{100}(15 + 1)=4 \). So the 4th value? Wait, no, wait. Wait, the locator method: if \( L_P \) is an integer, then the percentile is the average of the \( L_P \)-th and \( (L_P + 1) \)-th values? Wait, no, maybe I mixed up. Wait, let's recall: the 5 - number summary is minimum, Q1, median, Q3, maximum.

Minimum: smallest value, which is 31.

Maximum: largest value, which is 96.

Median: for \( n = 15 \) (odd), the median is the \( \frac{n + 1}{2}=\frac{16}{2}=8 \)-th term. So 8th term: let's list the data in order:

1: 31

2: 33

3: 34

4: 49

5: 53

6: 54

7: 55

8: 56

9: 64

10: 71

11: 72

12: 83

13: 89

14: 92

15: 96

So median (Q2) is 56 (8th term).

Now Q1: the median of the lower half. The lower half is the first 7 terms (since \( n = 15 \), lower half is positions 1 - 7: 31, 33, 34, 49, 53, 54, 55. The median of these 7 terms is the 4th term (since \( \frac{7 + 1}{2}=4 \)). So 4th term of lower half: 49. Wait, but using the locator method: \( L_{25}=\frac{25}{100}(15 + 1)=4 \). So the 4th term? Wait, the 4th term is 49. Let's check.

Q3: the median of the upper half. The upper half is positions 9 - 15: 64, 71, 72, 83, 89, 92, 96. The median of these 7 terms is the 4th term ( \( \frac{7 + 1}{2}=4 \) ). The 4th term of upper half is 83. Using locator method: \( L_{75}=\frac{75}{100}(15 + 1)=12 \). The 12th term? Wait, 12th term is 83. Let's check:

Wait, maybe the locator formula is \( L_P=\frac{P}{100}(n) \). Let's try that. For \( n = 15 \):

Minimum: 31 (1st term)

Q1: \( L_{25}=\frac{25}{100}\times15 = 3.75 \). Since it's not an integer, we round up to 4. So the 4th term: 49.

Median: \( L_{50}=\frac{50}{100}\times15 = 7.5 \). Round up to 8. 8th term: 56.

Q3: \( L_{75}=\frac{75}{100}\times15 = 11.25 \). Round up to 12. 12th term: 83.

Maximum: 96 (15th term).

Let's verify:

Data sorted: [31, 33, 34, 49, 53, 54, 55, 56, 64, 71, 72, 83, 89, 92, 96]

n = 15.

Minimum: 31.

Q1: The value at the 25th percentile. The formula for percentile: \( P_k = \text{value at the } \lceil k(n - 1)/100 ceil \)-th position? Wait, no, different methods. But the problem says to use the locator/percentile method in the book. Let's assume the locator is \( L = \frac{P}{100}(n + 1) \).

For Q1 (25th percentile): \( L = 0.25\times16 = 4 \). So the 4th term: 49.

For median (50th percentile): \( L = 0.5\times16 = 8 \). 8th term: 56.

For Q3 (75th percentile): \( L = 0.75\times16 = 12 \). 12th term: 83.

Maximum: 96.

Yes, that matches. So:

Minimum: 31

Q1: 49

Median: 56

Q3: 83

Maximum: 96

Answer:

31, 49, 56, 83, 96