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The times (in minutes) of a runner’s last 10 runs are 88, 65, 88, 49, 60, 71, 56, 87, 64, and 52. Find the standard deviation of the run times. Interpret your result.

Sagot :

Answer:

Standard deviation = 12.8 Approx.

Step-by-step explanation:

Given series;

88, 65, 88, 49, 60, 71, 56, 87, 64, and 52.

Find:

Standard deviation

Computation:

Assume mean (a) = 60

x        d(x - a)    d²

88        28       784

65         5        25

88        28       784

49        -11        121

60         0         0

71          11        121

56        -4         16

87        27        729

64         4         16

52        -8        64        

      Σd = 80   Σd² = 2,660

So,

Standard deviation = √[Σd² / N] - [Σd / N]²

Standard deviation = √[2260 / 10] - [80 / 10]²

Standard deviation = √226 - 64

Standard deviation = √162

Standard deviation = 12.8 Approx.

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