Question

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

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 2$

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 $X = 1.8, Z = -2$. We use this to find the new standard deviation $\sigma$. So

$Z = \frac{X - \mu}{\sigma}$

$-2 = \frac{1.8 - 2}{\sigma}$

$-2\sigma = -0.2$

$\sigma = \frac{0.2}{2}$

$\sigma = 0.1$

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