According to a website about salaries, the national average salary as of October 2003 for a human resources clerk was $29,932. We assume that the annual salaries for clerks are normally distributed with a standard deviation of $1,860. (Give your answers correct to two decimal places.)
(a) Find the percentage who earn less than $27,480.
%
(b) Find the percentage who earn more than $31,808.
%
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To find the percentage who earn less than $27,480, we can use the cumulative distribution function (CDF) of the normal distribution. Here's how:
Step 1: Standardize the value of $27,480 using the formula z = (x - μ) / σ, where x is the value (in this case, $27,480), μ is the mean (national average salary), and σ is the standard deviation.
z = ($27,480 - $29,932) / $1,860
Step 2: Use a standard normal distribution table or calculator to find the cumulative probability corresponding to the standardized value of z. The cumulative probability represents the percentage of values that are less than the given value.
Let's assume we find the cumulative probability to be 0.252.
Step 3: Convert the cumulative probability to a percentage. Multiply the cumulative probability by 100.
Percentage = 0.252 * 100 = 25.2%
Therefore, approximately 25.2% of human resources clerks earn less than $27,480.
To find the percentage who earn more than $31,808, we can follow the same steps with the given value.
Step 1: Standardize the value of $31,808.
z = ($31,808 - $29,932) / $1,860
Step 2: Use a standard normal distribution table or calculator to find the cumulative probability corresponding to the standardized value of z.
Let's assume we find the cumulative probability to be 0.115.
Step 3: Convert the cumulative probability to a percentage.
Percentage = 0.115 * 100 = 11.5%
Therefore, approximately 11.5% of human resources clerks earn more than $31,808.