Answer to Question #154683 in Statistics and Probability for Eugine Haweza

Question

1. An X-bar control chart monitors the mean of a process by checking that the average stays between and . When the process is under control,

(a) What is the probability that five consecutive sample means of n cases stay within these limits?

(b) What is the probability that all of the means for 100 days falls within the control limits?

Answer 1

Initial problem:

An X-bar control chart monitors the mean of a process by checking that the average stays between $\frac{μ-3σ}{\sqrt{n}}$ and $\frac{μ+3σ}{\sqrt{n}}$ . When the process is under control,

(a) What is the probability that five consecutive sample means of n cases stay within these limits?

(b) What is the probability that all of the means for 100 days falls within the control limits?

(a) The choice α = 0.0027 puts the control limits at $z_{α/2}=±3$ standard errors from the process target. The probability that a sample mean goes out of limit is 0.0027. This gives the probability that a sample mean stay within these limits is 1 - 0.0027 = 0.9973. Therefore, the probability that five consecutive sample means of n cases stay within these limits is:

$(0.9973)^5 = 0.986573$

The probability that five consecutive sample means of n cases stay within these limits is 0.986573.

(b) The choice α = 0.0027 puts the control limits at $z_{α/2}=±3$ standard errors from the process target. The probability that a sample mean goes out of limit is 0.0027. This gives the probability that a sample mean stay within these limits is 1 - 0.0027 = 0.9973. Therefore, the probability that all 100 sample means of 100 days (cases) stay within these limits is:

$(0.9973)^{100} = 0.763101$

The probability that all 100 sample means of 100 days stay within these limits is 0.763101.