Question

use the accompanying radiation levels (w/kg) for 50 different cell phones. find the percentile p25. p25 = \square w/kg (type an integer or a decimal. do not round.) \dots 0.24 0.28 0.35 0.53 0.55 0.56 0.62 0.66 0.78 0.88 0.89 0.92 0.93 0.94 0.95 0.95 0.96 1.00 1.05 1.07 1.08 1.10 1.11 1.12 1.12 1.14 1.19 1.19 1.20 1.20 1.22 1.22 1.24 1.26 1.26 1.27 1.30 1.30 1.31 1.32 1.35 1.36 1.39 1.44 1.46 1.46 1.47 1.48 1.52 1.53

Explanation:

Step1: Order the data

First, we need to order the given data set. Let's list out all the values (assuming we have 50 values, but let's check the visible ones and assume the rest are ordered as well. First, let's list the visible values: 0.24, 0.28, 0.35, 0.53, 0.55, 0.56, 0.62, 0.66, 0.78, 0.88, 0.89, 0.92, 0.93, 0.94, 0.95, 0.95, 0.96, 1.00, 1.05, 1.07, 1.08, 1.10, 1.11, 1.12, 1.12, 1.14, 1.19, 1.19, 1.20, 1.20, 1.22, 1.22, 1.24, 1.26, 1.26, 1.27, 1.30, 1.30, 1.31, 1.32, 1.35, 1.36, 1.39, 1.44, 1.46, 1.46, 1.47, 1.48, 1.52, 1.53. Wait, but we need 50 values. Maybe some are missing, but the process for percentile is:

The formula for the position of the \( p \)-th percentile (where \( p = 25 \) here) is \( i = \frac{p}{100} \times n \), where \( n \) is the number of data points. Here, \( n = 50 \), so \( i = \frac{25}{100} \times 50 = 12.5 \).

Since \( i \) is not an integer, we round up to the next integer, which is 13. So we need the 13th value in the ordered data set.

Wait, let's order the data properly. Let's list all the values (assuming the data is:

0.24, 0.28, 0.35, 0.53, 0.55, 0.56, 0.62, 0.66, 0.78, 0.88, 0.89, 0.92, 0.93, 0.94, 0.95, 0.95, 0.96, 1.00, 1.05, 1.07, 1.08, 1.10, 1.11, 1.12, 1.12, 1.14, 1.19, 1.19, 1.20, 1.20, 1.22, 1.22, 1.24, 1.26, 1.26, 1.27, 1.30, 1.30, 1.31, 1.32, 1.35, 1.36, 1.39, 1.44, 1.46, 1.46, 1.47, 1.48, 1.52, 1.53. Wait, that's 49 values? Maybe I missed one. Let's count again. Let's list each:

1. 0.24
2. 0.28
3. 0.35
4. 0.53
5. 0.55
6. 0.56
7. 0.62
8. 0.66
9. 0.78
10. 0.88
11. 0.89
12. 0.92
13. 0.93
14. 0.94
15. 0.95
16. 0.95
17. 0.96
18. 1.00
19. 1.05
20. 1.07
21. 1.08
22. 1.10
23. 1.11
24. 1.12
25. 1.12
26. 1.14
27. 1.19
28. 1.19
29. 1.20
30. 1.20
31. 1.22
32. 1.22
33. 1.24
34. 1.26
35. 1.26
36. 1.27
37. 1.30
38. 1.30
39. 1.31
40. 1.32
41. 1.35
42. 1.36
43. 1.39
44. 1.44
45. 1.46
46. 1.46
47. 1.47
48. 1.48
49. 1.52
50. 1.53 (Ah, I missed the 50th value earlier. So now \( n = 50 \))

Now, calculate the position \( i = \frac{25}{100} \times 50 = 12.5 \). Since \( i \) is a decimal, we take the average of the 12th and 13th values (wait, no: the rule for percentiles is: if \( i \) is not an integer, round up to the next integer. Wait, different sources have different methods. One common method is:

- If \( i \) is an integer, the percentile is the average of the \( i \)-th and \( (i+1) \)-th values.
- If \( i \) is not an integer, round up to the next integer, and the percentile is the value at that position.

Wait, let's check the formula for percentile:

The formula for the \( p \)-th percentile is:

1. Arrange the data in ascending order.
2. Calculate \( i = \frac{p}{100} \times n \), where \( n \) is the number of data points.
3. If \( i \) is an integer, the \( p \)-th percentile is the average of the \( i \)-th and \( (i+1) \)-th values.
4. If \( i \) is not an integer, round up to the next integer \( i' \), and the \( p \)-th percentile is the \( i' \)-th value.

So here, \( p = 25 \), \( n = 50 \), so \( i = \frac{25}{100} \times 50 = 12.5 \). Since \( i \) is not an integer, we round up to \( i' = 13 \).

Now, the 13th value in the ordered data set:

Let's list the ordered data with indices:

1: 0.24

2: 0.28

3: 0.35

4: 0.53

5: 0.55

6: 0.56

7: 0.62

8: 0.66

9: 0.78

10: 0.88

11: 0.89

12: 0.92

13: 0.93

Wait, but let's confirm the 12th value: index 12 is 0.92, index 13 is 0.93.

But wait, another method: some sources use linear interpolation. Let's see:

The formula for linear interpolation is:

\( P_p = x_{\lfloor i floor} + (i - \lfloor i floor)(x_{\lceil i ceil} - x_{\lfloor i floor}) \)

Here, \( i = 12.5 \), \( \lfloor i floor = 12 \), \( \lceil i ceil = 13 \)

\( x_{12} = 0.92 \), \( x_{13} = 0.93 \)

So \( P_{25} = 0.92 + (12.5 - 12)(0.93 - 0.92) = 0.92 + 0.5 \times 0.01 = 0.92 + 0.005 = 0.925 \)

Wait, but let's check the data again. Wait, maybe I ordered the data wrong. Let's check the original data:

The given data (from the image) has:

First column (top to bottom? Wait, the image shows a table with values. Let's re-express the data correctly. Let's list all the values as per the image:

Looking at the image, the data is:

0.24, 0.28, 0.35, 0.53, 0.55, 0.56, 0.62, 0.66, 0.78, 0.88,

0.89, 0.92, 0.93, 0.94, 0.95, 0.95, 0.96, 1.00, 1.05, 1.07,

1.08, 1.10, 1.11, 1.12, 1.12, 1.14, 1.19, 1.19, 1.20, 1.20,

1.22, 1.22, 1.24, 1.26, 1.26, 1.27, 1.30, 1.30, 1.31, 1.32,

1.35, 1.36, 1.39, 1.44, 1.46, 1.46, 1.47, 1.48, 1.52, 1.53

Yes, that's 50 values (10 rows, 5 columns? Wait, no, the first column has 10 values? Wait, no, let's count:

First set: 0.24, 0.28, 0.35, 0.53, 0.55, 0.56, 0.62, 0.66, 0.78, 0.88 (10)

Second set: 0.89, 0.92, 0.93, 0.94, 0.95, 0.95, 0.96, 1.00, 1.05, 1.07 (10)

Third set: 1.08, 1.10, 1.11, 1.12, 1.12, 1.14, 1.19, 1.19, 1.20, 1.20 (10)

Fourth set: 1.22, 1.22, 1.24, 1.26, 1.26, 1.27, 1.30, 1.30, 1.31, 1.32 (10)

Fifth set: 1.35, 1.36, 1.39, 1.44, 1.46, 1.46, 1.47, 1.48, 1.52, 1.53 (10)

Total: 10*5=50. Good.

Now, ordered data is correct as above.

Now, \( i = \frac{25}{100} \times 50 = 12.5 \)

Using linear interpolation:

\( P_{25} = x_{12} + (i - 12)(x_{13} - x_{12}) \)

\( x_{12} = 0.92 \) (12th value: index 12)

\( x_{13} = 0.93 \) (13th value: index 13)

\( i - 12 = 0.5 \)

So \( P_{25} = 0.92 + 0.5*(0.93 - 0.92) = 0.92 + 0.5*0.01 = 0.925 \)

Wait, but let's check the 12th and 13th values:

Index 1: 0.24

2: 0.28

3: 0.35

4: 0.53

5: 0.55

6: 0.56

7: 0.62

8: 0.66

9: 0.78

10: 0.88

11: 0.89

12: 0.92

13: 0.93

Yes, so the 12th value is 0.92, 13th is 0.93.

So using linear interpolation, the 25th percentile is 0.925.

Alternatively, some methods round up to the 13th value, which is 0.93, but the linear interpolation method is more accurate.

Wait, let's confirm the formula for percentile. The standard method (as per textbooks) is:

For a data set with \( n \) observations, ordered from smallest to largest, the \( p \)-th percentile is the value such that at least \( p\% \) of the observations are less than or equal to it and at least \( (100 - p)\% \) are greater than or equal to it.

The position \( i \) is calculated as \( i = \frac{p}{100} \times (n + 1) \) for some methods, but the most common is \( i = \frac{p}{100} \times n \).

Wait, let's check with \( i = \frac{25}{100} \times (50 + 1) = 12.75 \), which would also lead to interpolation. But the problem says "Do not round", so we need to use the correct method.

But in the problem, it's likely using the method where \( i = \frac{p}{100} \times n \), and if \( i \) is a decimal, we take the average of the \( \lfloor i floor \)-th and \( \lceil i ceil \)-th values.

So \( i = 12.5 \), so we take the average of the 12th and 13th values.

12th value: 0.92

13th value: 0.93

Average: \( \frac{0.92 + 0.93}{2} = \frac{1.85}{2} = 0.925 \)

So the 25th percentile \( P_{25} = 0.925 \)

Step2: Verify the position

We have 50 data points. 25% of 50 is 12.5, so the 25th percentile is between the 12th and 13th values. The 12th value is 0.92, the 13th is 0.93. Averaging them gives 0.925.

Answer:

0.925