Question

An article in Human Factors (June 1989) presented data on visual accommodation (a function of eye movement) when recognizing a speckle pattern on a high-resolution CRT screen. The data are as follows: 36.45, 67.90, 38.77, 42.18, 26.72, 50.77, 38.5, and 50.61. Calculate the sample mean and sample standard deviation. Round your answers to 2 decimal places. T43.89 Sample mean = || -12.35 Sample standard deviation =

Answer 1

We are given a data of sample size n = 8

36.45,67.90,38.77,42.18,26.72,50.77,38.5,50.61

Using this, first we find sample mean($\bar x$) and sample standard deviation(s).

$\bar x$ =   

sumofobservations

= (3.5 + 5.4.......+ 4.3)/5

$\bar x$ = 43.99

Now ,

s=   

I-U z(x - x) 3

Using given data, find Xi - $\bar x$ for each term.Take square for each.Then we can easily find s.

s= 12.41

Answer :

Sample mean :  43.99

Sample standard deviation:  12.41