ch 2.1 use the given frequency distribution to find the (a) class width, (b) class midpoints of the first…

ch 2.1 use the given frequency distribution to find the (a) class width, (b) class midpoints of the first class, (c) class boundaries of the first class. height (in inches) class frequency, f 50 - 52 5 53 - 55 8 56 - 58 12 59 - 61 13 62 - 64 11 4) short answer. write the word or phrase that best completes each statement or answers the question. 5) min = 1, max = 30, 6 classes use the maximum and minimum data entries and the number of classes to find the class width, the lower class limits, and the upper class limits. 5)

ch 2.1 use the given frequency distribution to find the (a) class width, (b) class midpoints of the first class, (c) class boundaries of the first class. height (in inches) class frequency, f 50 - 52 5 53 - 55 8 56 - 58 12 59 - 61 13 62 - 64 11 4) short answer. write the word or phrase that best completes each statement or answers the question. 5) min = 1, max = 30, 6 classes use the maximum and minimum data entries and the number of classes to find the class width, the lower class limits, and the upper class limits. 5)

Answer

Explanation:

Step1: Calculate class width

For the class 50 - 52, class width is found by subtracting the lower limit from the upper limit. So, $52 - 50=2$.

Step2: Calculate class mid - point

The formula for the mid - point of a class is $\frac{\text{lower limit}+\text{upper limit}}{2}$. For the class 50 - 52, it is $\frac{50 + 52}{2}=51$.

Step3: Calculate class boundaries

For a class with lower limit $L$ and upper limit $U$, the lower class boundary is $L-0.5$ and the upper class boundary is $U + 0.5$. For the class 50 - 52, the lower class boundary is $49.5$ and the upper class boundary is $52.5$.

Answer:

(a) 2 (b) 51 (c) 49.5 - 52.5