SOLUTION: Please help me solve this. An X-bar control chart monitors the mean of a process by checking that the average stays between control.pngand control2.png. When the process is und

Question 1172748: Please help me solve this.
An X-bar control chart monitors the mean of a process by checking that the average stays between control.pngand control2.png. 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?
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Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Absolutely! Let's break down this problem step by step.
**Understanding X-bar Control Charts**
An X-bar control chart monitors the mean of a process. The control limits (control.png and control2.png, which we'll refer to as UCL and LCL) are typically set at ±3 standard deviations from the process mean.
* **When the process is under control:** This means the process mean is stable, and the sample means should fall within the control limits with a high probability.
**Key Concepts**
* **Normal Distribution:** We assume the sample means follow a normal distribution.
* **Probability within Control Limits:** For a process under control, the probability that a single sample mean falls within the control limits is very high (close to 1). Specifically, within 3 standard deviations it is approximately 0.9973.
**Solving the Problem**
**(a) Probability of Five Consecutive Sample Means Within Limits**
1. **Probability for one sample mean:**
* Let P(in) be the probability that a single sample mean falls within the control limits.
* Since the process is under control, P(in) ≈ 0.9973 (for 3 sigma control limits).
2. **Probability for five consecutive sample means:**
* Since the samples are independent, we multiply the probabilities:
* P(5 consecutive in) = P(in) * P(in) * P(in) * P(in) * P(in) = P(in)^5
* P(5 consecutive in) = (0.9973)^5 ≈ 0.9866
* Therefore, the probability that five consecutive sample means stay within the limits is approximately 0.9866.
**(b) Probability of 100 Days of Means Within Limits**
1. **Probability for 100 days:**
* Similarly, for 100 days, we raise the probability for one day to the power of 100.
* P(100 consecutive in) = P(in)^100
* P(100 consecutive in) = (0.9973)^100 ≈ 0.7631
* Therefore, the probability that all of the means for 100 days fall within the control limits is approximately 0.7631.
**Summary**
* (a) The probability that five consecutive sample means stay within the limits is approximately 0.9866.
* (b) The probability that all of the means for 100 days fall within the control limits is approximately 0.7631.

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