Given the sample data. x: 27 14 25 22 31 (a) Find the range. (b) Verify that x =

Question

Given the sample data. x: 27 14 25 22 31

(a) Find the range.

(b) Verify that x = 119 and x2 = 2995. x x2

(c) Use the results of part (b) and appropriate computation formulas to compute the sample variance s2 and sample standard deviation s. (Enter your answers to one decimal place.) s2 s

(d) Use the defining formulas to compute the sample variance s2 and sample standard deviation s. (Enter your answers to one decimal place.) s2 s

(e) Suppose the given data comprise the entire population of all x values. Compute the population variance 2 and population standard deviation . (Enter your answers to one decimal place.) 2

Explanation / Answer

Solution

Part (a)

Range = maximum value – minimum value = 31 – 14 =17 ANSWER

Part (b)

x = 27 + 14 + 25 + 22 + 31 = 119

x2 = (272 + 142 + 252 + 222 + 312) = 729 + 196 + 625 + 484 + 961 = 2995

Part (c)

Sample variance, s2 = [x2 – {(x)2/n}]/(n - 1) = [2995 – {(119)2/5}]/4 = [2995 – 2832.2]/4

=162.8/4 = 40.7 ANSWER 1

Sample standard deviation. s = 40.7 = 6.38 = 6.4 ANSWER 2

Part (d)

Sample variance, s2 = [{(x – Xbar)2}/(n - 1)]

Now, Xbar = sample mean = (x)/n = 119/5 = 23.8

So, s2 = [(27 – 23.8)2 + (14 – 23.8)2 + (25 – 23.8)2 + (22 – 23.8)2 + (31 – 23.8)2]/4

= {3.22 + (-9.8)2 + 1.22 + (-1.8)2 + 7.22}/4 = (10.24 + 96.04 + 1.44 + 3.24 + 51.84)/4 = 162.8/4

= 40.7 ANSWER 1

Sample standard deviation. s = 40.7 = 6.38 = 6.4 ANSWER 2

Part (e)

Population variance, 2 = [{(x – µ)2}/n], where µ = population mean which, in this case (only), is the same as sample mean since sample is synonymous with population here.

So, 2 = [(27 – 23.8)2 + (14 – 23.8)2 + (25 – 23.8)2 + (22 – 23.8)2 + (31 – 23.8)2]/5

= {3.22 + (-9.8)2 + 1.22 + (-1.8)2 + 7.22}/5 = (10.24 + 96.04 + 1.44 + 3.24 + 51.84)/5 = 162.8/5

= 32.56 = 32. 6 ANSWER 1

and population standard deviation, = 32.56 = 5.71 = 5.7 ANSWER 2

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