Identify the class width, class midpoints, and class boundaries for the given frequency distribution.
daily low temperature (degrees°f) frequency daily low temperature (degrees°f) full data set frequency
45-48 1 61-64 749-52 3 65-68 753-56 5 69-72 157-60 11
a. what is the class width?
b. what are the class midpoints?
c. what are the class boundaries?

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

c.

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Intervalfrequency

45-481

49-523

53-565

57-6011

61-647

65-687

69-721

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

Class

IntervalMidpoints

45-48

49-52

53-56

57-60

61-64

65-68

69-72

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

IntervalClass boundary

45-4844.5-48.5

49-5248.5-52.5

53-5652.5-56.5

57-6056.5-60.5

61-6460.5-64.5

65-6864.5-68.5

69-7268.5-72.5

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.