Question

a chemical technician measured the temperatures (in °c) of 18 different solutions. the temperatures are shown below. complete the grouped relative frequency distribution for the data. (note that we are using a class width of 6.) write each relative frequency as a decimal rounded to the nearest hundredth, not as a percentage. temperature (in °c) 115 109 100 115 110 109 116 107 106 108 95 104 98 117 113 102 118 97 temperature (in °c) relative frequency 95 to 100 101 to 106 107 to 112 113 to 118

Explanation:

Step1: Count data in each class

- 95 to 100: Temperatures 95, 98, 97. So count = 3.
- 101 to 106: Temperatures 104, 106, 102. So count = 3.
- 107 to 112: Temperatures 107, 108, 109, 109, 110. So count = 5. Wait, wait, let's list all data: 115,109,100,115,110,109,116,107,106,108,95,104,98,117,113,102,118,97. Total 18 data points. Let's re - count:
- 95 - 100: 95, 98, 97, 100? Wait 100 is in 95 - 100? 95 ≤ x < 101 (since class width 6, 95 + 6 = 101). So 95, 98, 97, 100? Wait 100 is 100, which is less than 101. So 95,98,97,100? Wait original data: 95,98,97,100? Wait the data list: 115,109,100,115,110,109,116,107,106,108,95,104,98,117,113,102,118,97. Let's list each data and its class:
- 95: 95 - 100
- 109: 107 - 112 (107 ≤ 109 < 113)
- 100: 95 - 100
- 115: 113 - 118 (113 ≤ 115 < 119)
- 110: 107 - 112
- 109: 107 - 112
- 116: 113 - 118
- 107: 107 - 112
- 106: 101 - 106 (101 ≤ 106 < 107? Wait class width 6, 101+6 = 107. So 101 - 106: 101,102,103,104,105,106. So 104,106,102 are in 101 - 106.
- 108: 107 - 112
- 95: 95 - 100
- 104: 101 - 106
- 98: 95 - 100
- 117: 113 - 118
- 113: 113 - 118
- 102: 101 - 106
- 118: 113 - 118
- 97: 95 - 100

Now re - count:
- 95 - 100: 95, 100, 98, 97, 95? Wait no, 95,98,97,100,95? Wait no, the data points: 95,98,97,100, and is there another? Wait the data list: 95 (1), 98 (1), 97 (1), 100 (1). Wait 95 appears once, 98 once, 97 once, 100 once. Wait no, let's count again:
- 95 - 100: 95, 98, 97, 100, 95? Wait no, the data is [115,109,100,115,110,109,116,107,106,108,95,104,98,117,113,102,118,97]. So elements in 95 - 100 (95 ≤ x < 101): 95, 100, 98, 97, 95? Wait 95 is in the list once? Wait no, 95 is present once? Wait the data: 95 (index 10), 98 (index 12), 97 (index 17), 100 (index 2). So that's 4? Wait no, 95, 98, 97, 100. Wait 100 is 100, which is less than 101. So count = 4? Wait I made a mistake earlier. Let's list all data with their values:
- 95, 109, 100, 115, 110, 109, 116, 107, 106, 108, 95, 104, 98, 117, 113, 102, 118, 97. Now, for 95 - 100 (95 ≤ x < 101):
- 95 (yes), 100 (yes, 100 < 101), 98 (yes), 97 (yes), 95? Wait no, 95 is only once? Wait the data has 95 at position 10, 98 at 12, 97 at 17, 100 at 2. So that's 4 data points? Wait no, 95, 98, 97, 100: that's 4? Wait 95 is one, 98 one, 97 one, 100 one. So count = 4? Wait no, let's check the total number of data points. Let's count the list: 115 (1), 109 (2), 100 (3), 115 (4), 110 (5), 109 (6), 116 (7), 107 (8), 106 (9), 108 (10), 95 (11), 104 (12), 98 (13), 117 (14), 113 (15), 102 (16), 118 (17), 97 (18). So 18 data points. Now:
- 95 - 100: 95 (11), 100 (3), 98 (13), 97 (18). So count = 4.
- 101 - 106: 104 (12), 106 (9), 102 (16). So count = 3.
- 107 - 112: 107 (8), 108 (10), 109 (2), 109 (6), 110 (5). So count = 5.
- 113 - 118: 115 (1), 115 (4), 116 (7), 117 (14), 113 (15), 118 (17). Wait 115 is 115, which is in 113 - 118 (113 ≤ x < 119). So 115, 115, 116, 117, 113, 118. Count = 6. Wait but 4 + 3+5 + 6 = 18, which matches the total number of data points.

Step2: Calculate relative frequency (count / total)

- Total number of data points \( n = 18 \).
- For 95 to 100: Relative frequency \(=\frac{4}{18}\approx0.22\) (rounded to nearest hundredth).
- For 101 to 106: Relative frequency \(=\frac{3}{18}\approx0.17\) (rounded to nearest hundredth).
- For 107 to 112: Relative frequency \(=\frac{5}{18}\approx0.28\) (rounded to nearest hundredth).
- For 113 to 118: Relative frequency \(=\frac{6}{18} = 0.33\) (rounded to nearest hundredth). Wait, no: \(\frac{6}{18}=0.333\cdots\approx0.33\) when rounded to nearest hundredth. Wait but let's re - check the counts:
Wait, let's re - list the data for each class:
- 95 - 100: 95, 98, 97, 100 (4 values)
- 101 - 106: 102, 104, 106 (3 values)
- 107 - 112: 107, 108, 109, 109, 110 (5 values)
- 113 - 118: 113, 115, 115, 116, 117, 118 (6 values)
Now calculate relative frequencies:
- 95 - 100: \( \frac{4}{18}\approx0.22 \)
- 101 - 106: \( \frac{3}{18}\approx0.17 \)
- 107 - 112: \( \frac{5}{18}\approx0.28 \)
- 113 - 118: \( \frac{6}{18}=0.33 \) (since \( \frac{6}{18}=\frac{1}{3}\approx0.33 \) when rounded to the nearest hundredth)

Answer:

• 95 to 100: \( \boldsymbol{0.22} \)
• 101 to 106: \( \boldsymbol{0.17} \)
• 107 to 112: \( \boldsymbol{0.28} \)
• 113 to 118: \( \boldsymbol{0.33} \)