A population of plastic chairs in a factory has a weight's mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample of size 100 is selected and X is used to estimate u . What is the probability that the sample mean will be within +0.02 of the population mean ?

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 question, we have that:

What is the probability that the sample mean will be within +0.02 of the population mean?

Sample mean between 1.5 - 0.02 = 1.48 kg and 1.5 + 0.02 = 1.52 kg, which is the pvalue of Z when X = 1.52 subtracted by the pvalue of Z when X = 1.48. So

X = 1.52

By the Central Limit Theorem

has a pvalue of 0.9772

X = 1.48

has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

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step-by-step explanation:

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