Question

consider the following frequency table representing the scores on a test

class. Frequency

40-44. 3

45-49. 4

50-54. 12

55-59. 4

60-64. 15

determine the lower class boundary for the third class

Answer 1

Solution :

=> The distance from the upper limit of the first class to the lower limit of the second class is 45 - 44 = 1

=> Half this distance is = 0.5

=> The lower boundary of third class is calculated by subtracting half of the gap value from the third class lower limit

=> Third class lower boundary = 50 - 0.5 = 49.5