Question

In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes at 34 kV. The times, in minutes, are as follows: 0.19, 0.78, 0.96, 1.31, 2.78, 3.16, 4.15, 4.67, 4.85, 6.50, 7.35, 8.01, 8.27, 12.06, 31.75, 32.52, 33.91, 36.71, and 72.89. Calculate the sample mean and sample standard deviation.

Answer 1

Our data are:

0.19

0.78

0.96

1.31

2.78

3.16

4.15

4.67

4.85

6.50

7.35

8.01

8.27

12.06

31.75

32.52

33.91

36.71

72.89

From this data we have n=19

Mean is
`(\bar{x})` equal to:

`\bar{x} = (\sum x_i)/(n)`

We summarize all the values and divide them by the total number of samples.

`\bar{x}=(272.82)/(19)`

`\bar{x} = 14.3589`

In another hand we have that Standard Variation is equal to

`s=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n-1}}`

For this case we take the value of each of our samples, we subtract the average and we square it to that value. We do this with each of the data and in the end we divide them by the total population of the sample (n) minus 1.

`s= ((0.19-14.3589)^2.... (72.89-14.35)^2)/(19-1)`

`s=18.880`