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

Answer:

- the sample mean is 44

- the sample standard deviation is 12.35

Step-by-step 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 = $\sqrt{sum(x_{i} - xbar)^{2}/n-1}$

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

 = $\sqrt{\frac{1067.12}{7} }$

 = 12.35