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, 38.9, and 50.06. Calculate the sample mean and sample standard deviation.

Answer 1

Answer:

Mean = 43.969

Standard deviation = 12.341

Step-by-step explanation:

Given the data :

36.45, 67.90, 38.77, 42.18, 26.72, 50.77, 38.9, 50.06

The sample mean :

Σx / n = 351.75 / 8 = 43.969

Sample standard deviation :

√Σ(x - mean)²/n-1

√[(36.45-43.969)² + (67.90-43.969)² + (38.77-43.969)² + (42.18-43.969)² + (26.72-43.969)² + (50.77-43.969)² + (38.9-43.969)² + (50.06-43.969)² ] / ((8 - 1)

Standard deviation = 12.341