What's the standard form for 2.73x10^5
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Answer:
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 80 employees will be at most $75 is 0.9297.
Step-by-step explanation:
According to the Central Limit Theorem if we have a non-normal population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample means is given by,
And the standard deviation of the distribution of sample means is given by,
The information provided is:
As <em>n</em> = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.
Let represent the sample mean weekly salaries.
The distribution of is:
Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.
The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, .
Compute the probability as follows:
Thus, 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 80 employees will be at most $75 is 0.9297.