How frequently would two consecutive controls being beyond -2s from the mean occur on the basis of chance alone?

How frequently would two consecutive controls being beyond -2s from the mean occur on the basis of chance alone?

• A. 1:100
• B. 5:100
• C. 1:400
• D. 1:1,600

Answer: (D)

When considering the frequency of two consecutive control measurements falling outside -2 standard deviations (s) from the mean, we can reference the properties of the normal distribution, often applied in quality control. In a normal distribution, approximately 95% of the values will fall within ±2 standard deviations from the mean. This means that about 5% of the values, or 1 in 20, will fall outside this range in either tail (both below -2s and above +2s). Specifically, the probability of a control measurement being beyond -2s is about 2.5%, or 0.025. When looking at two consecutive control measurements, we consider their probabilities independently. The probability of both being beyond -2s from the mean would thus be calculated by multiplying the probability of the first event (0.025) by the probability of the second event (0.025): 0.025 x 0.025 = 0.000625. To express this in terms of occurrence per 100 (or 1 in X), we take the reciprocal of the probability: 1 / 0.000625 = 1600. Therefore, the chance of having two consecutive controls outside -2s from the mean occurs at

When considering the frequency of two consecutive control measurements falling outside -2 standard deviations (s) from the mean, we can reference the properties of the normal distribution, often applied in quality control.

In a normal distribution, approximately 95% of the values will fall within ±2 standard deviations from the mean. This means that about 5% of the values, or 1 in 20, will fall outside this range in either tail (both below -2s and above +2s). Specifically, the probability of a control measurement being beyond -2s is about 2.5%, or 0.025.

When looking at two consecutive control measurements, we consider their probabilities independently. The probability of both being beyond -2s from the mean would thus be calculated by multiplying the probability of the first event (0.025) by the probability of the second event (0.025):

0.025 x 0.025 = 0.000625.

To express this in terms of occurrence per 100 (or 1 in X), we take the reciprocal of the probability:

1 / 0.000625 = 1600.

Therefore, the chance of having two consecutive controls outside -2s from the mean occurs at