Temperature (°F)Frequency Use the given frequency distribution to find the (a) c

Question

Temperature (°F)Frequency Use the given frequency distribution to find the (a) class width. (b) class midpoints. (c) class boundaries 35-39 40-44 45-49 50-54 55-59 60-64 65-69 3 5 7 7 (a) What is the class width? (Type an integer or a decimal.) (b) What are the class midpoints? compiete the table below. Type integers or decimals) Temperature F/FrequencyMidpoint 35-39 40-44 45-49 50-54 55-59 60-64 65-69 3 (c) What are the class boundaries? Complete the table below. (Type integers or decimals.) Temperature CF)Frequenoyl claess bouncares 35-39 40-44 45-49 50-54 55-59 60-64 65-69 3 5

Explanation / Answer

Solution :

(a)

Class width is the difference between the upper or lower class limits of consecutive classes .

Width = 40 - 35 = 5

(b)

Class midpoints :

Temperature (0 F) Frequency Midpoint

35 - 39 1 (35 + 39) / 2 = 37

40 - 44 3 (40 + 44) / 2 = 42

45 - 49 5 (45 + 49) / 2 = 47

50 - 54 11 (50 + 54) / 2 = 52

55 - 59 7 (55 + 59) / 2 = 57

60 - 64 7 (60 + 64) / 2 = 62

65 - 69 1 (65 + 69) / 2 = 67

(c)

Class boundaries :

Gap = 40 - 39 = 1

Add, 1 / 2 = 0.5

Temperature (0 F) Frequency Lower boudaries Upper boundaries

35 - 39 1 35 - 0.5 = 34.5 39 + 0.5 = 39.5

40 - 44 3 40 - 0.5 = 39.5 44 + 0.5 = 44.5

45 - 49 5 45 - 0.5 = 44.5 49 + 0.5 = 49.5

50 - 54 11 50 - 0.5 = 49.5 54 + 0.5 = 54.5

55 - 59 7 55 - 0.5 = 54.5 59 + 0.5 = 59.5

60 - 64 7 60 - 0.5 = 59.5 64 + 0.5 = 64.5

65 - 69 1 65 - 0.5 = 64.5 69 + 0.5 = 69.5

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