The distribution of weekly salaries at a large company is right skewed with a mean of $1000 and a standard deviation of $350. What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation, which is also called standard error

In this problem, we have that:

What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

This is the pvalue of Z when X = 1000+50 = 1050 subtracted by the pvalue of Z when X = 1000-50 = 950. So

X = 1050

By the Central Limit Theorem

has a pvalue of 0.8438

X = 950

has a pvalue of 0.1562

0.8438 - 0.1562 = 0.6876

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50