A wire bonding process is said to be "in control" if the mean pull strength is 10 pounds. It is known that pull-strength measurements are normally distributed with a standard deviation of 1.5 pounds. Periodic random samples of size 4 are taken from this process and the process is said to be "out of control" if a sample mean is less than 8.25 pounds.
1. What is the probability that the process will be declared "out of control" when the true mean pull strength is 10 pounds? A. 0.0021
B. 0.0099
C. 0.0548
D. 0.1217

B. 0.0099

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this roblem, we have that:

Probability of being out of control.

Less than 8.25 pounds, which is the pvalue of Z when X = 8.25. So

By the Central Limit Theorem

has a pvalue of 0.0099.

So the correct answer is:

B. 0.0099

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step-by-step explanation:

i think it is 18

step-by-step explanation: