Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 1000, \sigma = 350, n = 50, s = \frac{350}{\sqrt{50}} = 49.5[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1050-1000}{49.5}[/tex]
[tex]Z = 1.01[/tex]
[tex]Z = 1.01[/tex] has a pvalue of 0.8438
X = 950
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{950-1000}{49.5}[/tex]
[tex]Z = -1.01[/tex]
[tex]Z = -1.01[/tex] 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
