A manufacturer orders lots of steel rods from a supplier (each lot consists of thousands of rods). Assume that the mean length of the rods in each lot is 4 meters and standard deviation is 0.04 meter. The quality manager accepts a lot if the mean length of the randomly selected sample of 64 rods from that lot is at least 4.005 meters, otherwise the lot is returned to the supplier. a) What percentage of the lots received by the manufacturer will be returned to the supplier

84.13% of the lots received by the manufacturer will be returned to the supplier

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 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.

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.

Assume that the mean length of the rods in each lot is 4 meters and standard deviation is 0.04 meter.

This means that

Sample of 64:

This means that

a) What percentage of the lots received by the manufacturer will be returned to the supplier

The proportion is the pvalue of Z when X = 4.005. So

By the Central Limit Theorem

has a pvalue of 0.8413

0.8413*100% = 84.13%

84.13% of the lots received by the manufacturer will be returned to the supplier

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step-by-step explanation: