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)
95 108 112 113 117
104 118 107 110 116
109 100 104 99 103
115 96 114

temperature (in °c)\trelative frequency
95 to 100\t
101 to 106\t
107 to 112\t
113 to 118\t

Explanation:

Step1: Count data in each class

• For 95 - 100: The values in this range are 95, 99, 100, 96. So the frequency $f_1 = 4$.
• For 101 - 106: The values are 104, 104, 103. So the frequency $f_2=3$.
• For 107 - 112: The values are 112, 107, 110, 109. So the frequency $f_3 = 4$.
• For 113 - 118: The values are 113, 117, 118, 116, 115, 114. So the frequency $f_4=6$.

Step2: Calculate total frequency

The total number of data points $n=18$.

Step3: Calculate relative - frequencies

• The relative frequency formula is $rf=\frac{f}{n}$.
• For 95 - 100: $rf_1=\frac{f_1}{n}=\frac{4}{18}\approx0.22$.
• For 101 - 106: $rf_2=\frac{f_2}{n}=\frac{3}{18}\approx0.17$.
• For 107 - 112: $rf_3=\frac{f_3}{n}=\frac{4}{18}\approx0.22$.
• For 113 - 118: $rf_4=\frac{f_4}{n}=\frac{6}{18}\approx0.33$.

Answer:

Temperature (in °C)	Relative frequency
101 to 106	0.17
107 to 112	0.22
113 to 118	0.33