(a) 137and −354 (b) − 52 and 52
(c) −312, 39 and 192 (d) − 50, − 200 and 300
(a) 137 +(−354) = −217
(b) −52+ 52 = 0
(c) −312+ 39 + 192 = −312 + 231 = −81
(d) −50+ (−200) + 300 = −250 + 300 = 50
(a) 137 and −354 (b) − 52 and 52
(c) − 312, 39 and 192 (d) − 50, − 200 and 300
There is a grouped data distribution for which mean is to be found by step deviation method.
Class IntervalNumber of Frequency(fi)Class mark(xi)di=xi−aui=dih0−1004050−200D......100−20039150B.....E.....200−3003425000300−400303501001400−50045450C.....F......TotalA=∑fi=......
Find the value of A, B, C, D, E and F respectively.
Find the sum of:
–50,–200 and 300