Question

A student union cafeteria worker checked the weight of ten half-pound bags of whole bean coffee and recorded the following weights in pounds: 0.48, 0.51, 0.47, 0.49, 0.49, 0.50, 0.52, 0.48, 0.49, 0.51. What is the standard deviation of the weight of these coffee bags?

Answer 1

Answer: σ = 0.0154

Step-by-step explanation:

1. Calculate the mean for the recorded data.

(0.48 + 0.51 + 0.47 + 0.49 + 0.49 + 0.50 + 0.52 + 0.48 + 0.49 + 0.51)/10 =

4.94 / 10 = 0.49

The mean is represented with the greek letter μ, so μ = 0.49

2. For each element in the serie, subtract the mean (0.49) and square the result

(0.48 - 0.49)^2 = (-0.01)^2 = 0.0001

(0.51 - 0.49)^2 = (0.02)^2 = 0.0004

(0.47 - 0.49)^2 = (-0.02)^2 = 0.0004

(0.49 - 0.49)^2 = 0

(0.49 - 0.49)^2 = 0

(0.50 - 0.49)^2 = (0.01)^2 = 0.0001

(0.52 - 0.49)^2 = (0.03)^2 = 0.0009

(0.48 - 0.49)^2 = (-0.01)^2 = 0.0001

(0.49 - 0.49)^2 = 0

(0.51 - 0.49)^2 = (0.02)^2 = 0.0004

So we get these results: 0.0001, 0.0004, 0.0004, 0, 0, 0.0001, 0.0009, 0.0001, 0, 0.0004

3. Now, we calculate the mean of these squared differences

(0.0001 + 0.0004 + 0.0004 + 0 + 0 + 0.0001 + 0.0009 + 0.0001 + 0 + 0.0004) / 10 =

(0.0024) / 10 = 0.00024

4. Now to get the standard deviation, we need to get the root of this last value (0.00024).

√(0.00024) = 0.0154

5. So that will be our standard deviation value:

σ = 0.0154

DONE!