Temperature (F) 40-44 45-49 50-54 55-59 60-64 65-69 70-74 Use the given frequenc

Question

Temperature (F) 40-44 45-49 50-54 55-59 60-64 65-69 70-74 Use the given frequency distribution to find the (a) class width (b) class midpoints (c) class boundaries Frequency (a) What is the class width? (Type an integer or a decimal.) (b) What are the class midpoints? Complete the table below. (Type integers or decimals.) Temperature (F) Frequency Midpoint 40-44 45-49 50-54 55-59 60-64 65-69 70-74 (c) What are the class boundaries? Complete the table below. (Type integers or decimals.) Temperature (%) 40-44 45-49 50-54 55-59 60-64 65-69 70-74 Frequency Class boundaries

Explanation / Answer

a) The class width is the difference between the upper or lower class limits of consecutive classes. All classes should have the same class width. In this case,class width equals to the difference between the lower limits of the first two classes.

So, class width = 45 - 40 = 5

b) The class midpoint is the lower class limit plus the upper class limit divided by 2.

So, here

class midpoints:

(40 + 44)/2 = 42

(45 + 49)/2 = 47

(50 + 54)/2 = 52

(55 + 59)/2 = 57

(60 + 64)/2 = 62

(65 + 69)/2 = 67

(70 + 74)/2 = 72

c) On the other hand, class boundaries were called true class limits. These are boundary points. They tend to bound the class limits.

Here's the definitional formula for class boundaries

UCB = UL+1/2 unit of measurement

LCB = LL12 unit of measurement

i) UCB = 44 + (1/2) * 1 = 44.5

LCB = 40 - (1/2) * 1 = 39.5

So, class boundary = 39.5 - 44.5

ii) 44.5 - 49.5

iii) 49.5 - 54.5

iv) 54.5 - 59.5

v) 59.5 - 64.5

vi) 64.5 - 69.5

vii) 69.5 - 74.5

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