Introduction:

The empirical rule for a normally distributed random variable is as follows:

About \(68\%\) of the data lie within the 1-standard deviation interval about the mean,

About \(95\%\) of the data lie within the 2-standard deviation interval about the mean,

About \(99.7\%\) of the data lie within the 3-standard deviation interval about the mean.

Explanation:

When the process is still in control, the sample mean can be said to follow the normal distribution with the true value of the characteristic of interest as the mean (center line), and the relevant standard error of sample mean. In this case, the above empirical rule will hold true for the process.

Calculation:

Assume \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) to be three consecutive sample means obtained from the process. Since the samples are all obtained at random, the values taken by the sample mean will be independent of one another. Further, being taken from the same process, \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) are identically distributed.

Now, it is given that each of the three means are above the +1-standard deviation line, that is, above the upper bound of the 1-standard deviation interval of the mean. Consider the case of \(\displaystyle{X}_{{1}}\).

Since \(68\%\) of the data lie within the 1-standard deviation interval, the remaining \(\displaystyle{32}\%{\left(={100}\%–{68}\%\right)}\) lie outside the 1-standard deviation interval. The normal distribution being symmetric, half of this \(32\%\) would lie above the upper bound of the 1-standard deviation interval, while the remaining half would lie below the lower bound of the interval.

Thus, \(\displaystyle{X}_{{1}}\) would lie above the upper bound of the 1-standard deviation interval, that is, above the +1-standard deviation line, with a \(\displaystyle{16}\%{\left(={32}\%\text{/}{2}\right)}\) or 0.16 probability.

Since, \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) are identically distributed, each of bar \(\displaystyle{X}_{{2}}{\quad\text{and}\quad}\overline{{X}}_{{3}}\) would lie above the +1-standard deviation line with probability 0.16.

The distributions of \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) being independent, the probability that all three of bar \(\displaystyle{X}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) lie above the +1-standard deviation line is the product of the individual probabilities, that is, \(\displaystyle{\left({0.16}\right)}\cdot{\left({0.16}\right)}\cdot{\left({0.16}\right)}={0.004096}.\)

Hence, if the process is still in control, the probability that 3 sample means in a row would plot above the +1-standard deviation line is 0.004096.

The empirical rule for a normally distributed random variable is as follows:

About \(68\%\) of the data lie within the 1-standard deviation interval about the mean,

About \(95\%\) of the data lie within the 2-standard deviation interval about the mean,

About \(99.7\%\) of the data lie within the 3-standard deviation interval about the mean.

Explanation:

When the process is still in control, the sample mean can be said to follow the normal distribution with the true value of the characteristic of interest as the mean (center line), and the relevant standard error of sample mean. In this case, the above empirical rule will hold true for the process.

Calculation:

Assume \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) to be three consecutive sample means obtained from the process. Since the samples are all obtained at random, the values taken by the sample mean will be independent of one another. Further, being taken from the same process, \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) are identically distributed.

Now, it is given that each of the three means are above the +1-standard deviation line, that is, above the upper bound of the 1-standard deviation interval of the mean. Consider the case of \(\displaystyle{X}_{{1}}\).

Since \(68\%\) of the data lie within the 1-standard deviation interval, the remaining \(\displaystyle{32}\%{\left(={100}\%–{68}\%\right)}\) lie outside the 1-standard deviation interval. The normal distribution being symmetric, half of this \(32\%\) would lie above the upper bound of the 1-standard deviation interval, while the remaining half would lie below the lower bound of the interval.

Thus, \(\displaystyle{X}_{{1}}\) would lie above the upper bound of the 1-standard deviation interval, that is, above the +1-standard deviation line, with a \(\displaystyle{16}\%{\left(={32}\%\text{/}{2}\right)}\) or 0.16 probability.

Since, \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) are identically distributed, each of bar \(\displaystyle{X}_{{2}}{\quad\text{and}\quad}\overline{{X}}_{{3}}\) would lie above the +1-standard deviation line with probability 0.16.

The distributions of \(\displaystyle\overline{{X}}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) being independent, the probability that all three of bar \(\displaystyle{X}_{{1}},\overline{{X}}_{{2}},\overline{{X}}_{{3}}\) lie above the +1-standard deviation line is the product of the individual probabilities, that is, \(\displaystyle{\left({0.16}\right)}\cdot{\left({0.16}\right)}\cdot{\left({0.16}\right)}={0.004096}.\)

Hence, if the process is still in control, the probability that 3 sample means in a row would plot above the +1-standard deviation line is 0.004096.