The product of c and 2
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Answer:
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