Questions: A chemical technician measured the temperatures (in °C) of 13 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 4.) Write each relative frequency as a decimal rounded to the nearest hundredth, not as a percentage. Temperature (in °C) 99 94 103 97 109 108 95 100 93 104 102 101 102 Temperature (in °C) Relative frequency 93 to 96 97 to 100 101 to 104 105 to 108 109 to 112

Solution Steps

To complete the grouped relative frequency distribution, we need to count how many temperatures fall into each specified range (class). Then, we calculate the relative frequency by dividing the count for each class by the total number of temperatures. Finally, we round the relative frequencies to the nearest hundredth.

The class intervals are defined as follows:

• $93 \leq T < 97$
• $97 \leq T < 101$
• $101 \leq T < 105$
• $105 \leq T < 109$
• $109 \leq T < 113$

Count the number of temperatures in each class interval:

• $93 \leq T < 97$: 3 temperatures (94, 95, 93)
• $97 \leq T < 101$: 3 temperatures (99, 97, 100)
• $101 \leq T < 105$: 5 temperatures (103, 104, 102, 101, 102)
• $105 \leq T < 109$: 1 temperature (108)
• $109 \leq T < 113$: 1 temperature (109)

Calculate the relative frequency for each class interval by dividing the count by the total number of temperatures, 13:

• $93 \leq T < 97$: $\frac{3}{13} \approx 0.23$
• $97 \leq T < 101$: $\frac{3}{13} \approx 0.23$
• $101 \leq T < 105$: $\frac{5}{13} \approx 0.38$
• $105 \leq T < 109$: $\frac{1}{13} \approx 0.08$
• $109 \leq T < 113$: $\frac{1}{13} \approx 0.08$

Final Answer