Question

The standard devision of the 5,15,20,25,30,35,45,55 lies between​

Answer 1

Answer:

Workout :

step 1 Address the formula, input parameters & values

Input parameters & values

x1 = 5, x2 = 10, . . . . , x10 = 50

Number of samples n = 10

Degrees of freedom = n - 1

= (10 - 1) = 9

Find sample standard deviation of 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50

step 2 Find the mean for 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50

µ =

n

∑

i = 1

Xi

n

=(5 + 10 + 15 + . . . . + 50)

10

µ = 27.5

step 3 Apply the samples, mean & degrees of freedom values in the below sample standard deviation formula

=√{ (5 - 27.5)² + (10 - 27.5)² + (15 - 27.5)² + . . . . + (50 - 27.5)²}/9

=√(-22.5)² + (-17.5)² + (-12.5)² + . . . . + (22.5)²

9

=√(506.25 + 306.25 + 156.25 + . . . . + 506.25)

9

=√2062.5

9

= √229.1667

s = 15.1383

For the dataset 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50

27.5 is the mean

15.1383 σ is the estimated (sample) standard deviation

229.1667 is estimated (sample) variance

14.3614 σ is the standard deviation of finite population

206.25 is the variance of finite population

Step-by-step explanation:

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