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