Question

6.1.7 (Video Solution) 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, 39.0, and 50.23. Calculate the sample mean and sample standard deviation. Round your answers to 2 decimal places.

Answer 1

- the sample mean is 44

- the sample standard deviation is 12.35

Explanation:

given information:

data,
`x_(i)` = 36.45, 67.90, 38.77, 42.18, 26.72, 50.77, 39.0, 50.23

the number of data, n = 8

the sample mean, xbar

xbar = ∑
`x_(i)`/n

= (36.45+67.90+38.77+42.18+26.72+50.77+39.0+50.23)/8

= 352.08/8

= 44

standard deviation, s

s =
`\sqrt{sum(x_(i) - xbar)^(2)/n-1}`

=
`\sqrt{(36.45-44)^(2)+(36.45-44)^(2).........(39.00-44)^(2)+(50.23-44)^(2)/(8-1 )}`

=
`\sqrt{(1067.12)/(7) }`

= 12.35