Question

class boundaries	midpoint	frequencies
16.5-23.5		5
23.5-30.5		6
30.5-37.5		9
37.5-44.5		7
44.5-51.5		3

1) Find the class midpoints

2)  Mean

3) Standard deviation

4) Variance

Answer 1

Solution:

Class (1)	Frequency (f) (2)	Mid value (x) (3)	f⋅x (4)=(2)×(3)	f⋅x2=(f⋅x)×(x) (5)=(4)×(3)
16.5-23.5	5	20	100	2000
23.5-30.5	6	27	162	4374
30.5-37.5	9	34	306	10404
37.5-44.5	7	41	287	11767
44.5-51.5	3	48	144	6912
	n=30		∑f⋅x=999	∑f⋅x2=35457

Mean ˉx=∑fxn

=99/30

=33.3

Sample Variance S2=∑f⋅x2-(∑f⋅x)2nn-1

=35457-(999)230/29

=35457-33266.7/29

=2190.329

=75.5276


Sample Standard deviation S=√∑f⋅x2-(∑f⋅x)2nn-1

=√35457-(999)230/29

=√35457-33266.7/29

=√2190.3/29

=√75.5276

=8.690