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?
The probabilities are based on a normal (not skewed) distribution. Calculations will be in error, depending on the degree of skewness.
To find the probability that the sampling error will be at most $50, we need to calculate the z-score and then find the corresponding probability from the standard normal distribution.
The formula for calculating the z-score is:
z = (X - μ) / (σ / sqrt(n))
Where:
- X is the value we want to find the probability for (in this case, $50)
- μ is the population mean ($1000)
- σ is the population standard deviation ($350)
- n is the sample size (50)
Substituting these values into the formula:
z = (50 - 1000) / (350 / sqrt(50))
Calculating the z-score:
z = -950 / (350 / 7.07)
z = -950 / 49.50
z = -19.19
Now we need to find the probability corresponding to this z-score. Using a Z-table or a calculator, we can find that the probability associated with a z-score of -19.19 is extremely close to 0.
Therefore, 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 50 employees will be at most $50 is practically 0.
To find the probability that the sampling error is at most $50, we need to calculate the probability that the absolute difference between the sample mean and the population mean is less than or equal to $50.
The sampling error is defined as the difference between the sample mean and the population mean. In this case, the mean of the random sample of weekly salaries is the sample mean, and the population mean is given as $1000.
The formula to calculate the sampling error is:
Sampling error = Sample mean - Population mean
Given that the distribution of weekly salaries is right-skewed with a mean of $1000 and a standard deviation of $350, we can assume that the sample mean will be normally distributed due to the Central Limit Theorem, as the sample size is reasonably large (n > 30).
To find the probability, we need to standardize the sampling error using z-scores and then consult a standard normal distribution table or calculator.
The formula to calculate the z-score is:
z = (Sampling error - 0) / (Standard deviation / sqrt(sample size))
In this case:
Sampling error = $50
Standard deviation = $350
Sample size = 50
Plugging in the values:
z = (50 - 0) / (350 / sqrt(50))
Using a calculator, we can calculate that z ≈ 0.5352.
Next, we need to find the area under the standard normal curve to the left of the z-score of 0.5352. This gives us the probability that the sampling error is less than or equal to $50.
Using a standard normal distribution table or a calculator, we find that P(Z ≤ 0.5352) ≈ 0.7031.
Therefore, 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 50 employees will be at most $50 is approximately 0.7031, or 70.31%.