Question

The following data represent glucose blood levels (mg/100 ml) after a 12-hour fast for a random sample of 70 women (Reference American Journal of Clinical Nutrition, Vol. 19, pp. 345-351) 45 668) 71 75 64 59 59 75 82 B0 81 85 77 82 90 87 72 70 69 83 71 87 69 81 76 96 83 67 94 101 94 89 94 73 99 93 85 83 90 78 80 85 83 84 74 81 70 65 89 70 80 84 77 55 46 80 70 75 45 101 71 109 73 73 80 72 81 63 74 For this problem, use six classes (a) Find the class width (b) Make a frequency table showing class limits, class boundaries, midpoints, frequencies, relative frequencies, and cumulative frequencies. (Give relative frequencies to 4 decimal places.) Relative Cumulative Class Boundaries Midpoint Frequency Frequency Frequency (c) Draw a histogram. 555657585 5555657585 95 105

Answer 1

(a) In the above question, we have given to use 6 bins for deciding class width

then the class width should be of 11 that is the difference between the upper-class limit and lower class limit

like 44.5, 55.5, 66.5, 77.5, 88.5, 99.5, 110.5 as an exclusive interval in which upper limit is excluded for that interval.

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(b)  

Class Limits	Class Boundaries	Midpoint	Frequency	Relative frequency	Cumulative frequency
44-55	44.5-55.5	49.5	3	3/70 = 0.0428	3
56-67	55.5-66.5	61.5	8	8/70 = 0.1142	3+8 = 11
68-79	66.5-77.5	73.5	23	23/70 = 0.3285	11+23 = 34
80-91	77.5-88.5	85.5	27	27/70 = 0.3857	34+27 = 61
92-103	88.5-99.5	97.5	8	8/70 = 0.1142	61+8 = 69
104-115	99.5-110.5	109.5	1	1/70 = 0.0142	69+1 = 70

$\boxed{Midpointpoint = \frac{Upper~class~limit~+~lower~class~limit}{2}}$, formula used

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(C) In the given histogram there is only one correct histogram is givem which is 4th because class interval of 1st histogram is not correct , 2nd and 3rd hisstograms do not have correct frequencies.

Hence, the 4th histogram is the correct histogram.

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