A production manager at a wall clock company wants to test their new wall clocks. The designer claims they have a mean life of 14 years with a variance of 25.
If the claim is true, in a sample of 50 wall clocks, what is the probability that the mean clock life would be greater than 12.5 years? Round your answer to four decimal
places.

0.9830 = 98.30% probability that the mean clock life would be greater than 12.5 years.

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(square root of the variance) , 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 problem, we have that:

What is the probability that the mean clock life would be greater than 12.5 years?

This is 1 subtracted by the pvalue of Z when X = 12.5. So

has a pvalue of 0.0170.

1 - 0.0170 = 0.9830

0.9830 = 98.30% probability that the mean clock life would be greater than 12.5 years.

there is no graph so i cant sorry

step-by-step explanation:

parallelogram is the   answer