A bottle-filling process has a low specification limit of 1.8 liter and upper specification limit of 2.2 liter. The standard deviation is 0.15 liter, and the mean is 2 liter. The company now wants to reduce its defect probability as 0.0455. To what level would it have to reduce the standard deviation in the process to meet this target

The standard deviation would have to be reduced to 0.1 in the process to meet this target

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The mean is 2 liter.

This means that

The company now wants to reduce its defect probability as 0.0455.

This means that:

P(X < 1.8) = 0.0455/2 = 0.02275

P(X < 2.2) = 0.0455/2 = 0.02275

This means that the pvalue of Z when X = 1.8 is 0.02275. This means that when . We use this to find the new standard deviation . So

The standard deviation would have to be reduced to 0.1 in the process to meet this target