Question

A person measures the time period of 50 oscillations
of a simple pendulum 5 times. The data set is 46 s.
44.5 S, 47.5 s, 45 s and 47 s. If least count of the
measuring clock is 0.5 s, then the reported time
period should be
(1) (46 -+1.4) (2) (46 + 1)
(3) (46 + 2) s . (4) (46 1.5) s'​

Answer 1

Answer:

(46 +- 1).

Explanation:

Since, there are 5 observation given while the person measures the time period. So, the arithmetic mean of the time period will be xbar=∑xi/n where n is the number of observation. So, xbar=(46+44.5+47.5+45+47)/5 which on solving we will get 46.

While the standard mean deviation will be ∑|xbar - xi|/n = (0+1.5+1.5+1+1)/5 which on solving we will get +-1. So, the reported time period will be (46 +- 1).

Answer 2

Answer:

46 ± 1

Explanation:

The data set is 46 s. 44.5 S, 47.5 s, 45 s and 47 s

Mean = ( 46 + 44.5 + 47.5 + 45 + 47) /5

= 230/5

= 46

|46 - 46|  =   0

| 44.5 - 46| = 1.5

|47.5 - 46| = 1.5

|45 - 46| = 1

|47 - 46| = 1

Mean = (0 + 1.5 + 1.5 + 1 + 1)/5

=5/5

= 1

Reported time Should be  46 ± 1