Question

The BLS employer cost survey uses a sample to establish the average wage of receptionists. Based on a large number of observations, the distribution of receptionist wages is normally distributed with a mean $10.38/hour and a standard deviation of $2.05. What is the probability that the wages for a sample of 20 receptionists exceeds $11/hour? Multiple Choice 0.3000 0.0881 0.0222 0.5498

Answer 1

To find the probability that the wages for a sample of 20 receptionists exceeds $11/hour, calculate the standard error of the mean, determine the z-score for $11, and use the standard normal distribution to find the probability. The correct answer is 0.0222.

Step-by-step explanation:

The question you're asking relates to the concept of sampling distributions and probabilities in statistics. When talking about the mean wage of a sample of receptionists, we're dealing with a sample mean, which is a random variable. As the sample size increases, according to the Central Limit Theorem, the distribution of the sample mean will approach a normal distribution, even if the population distribution is not exactly normal. This means we can use z-scores to find the probability you're looking for.

Here, the population mean (μ) is $10.38/hour and population standard deviation (σ) is $2.05/hour. To determine the probability that the mean wage for a sample of 20 receptionists exceeds $11/hour, you first need to calculate the standard error of the mean (SEM), which is σ divided by the square root of the sample size (n).

SEM = σ / √ n = $2.05 / √ 20

Then, compute the z-score for $11, which is the sample mean minus the population mean divided by the SEM:

Z = ($11 - $10.38) / SEM

The z-score tells us how many standard deviations the sample mean is away from the population mean. Once we have the z-score, we can use the standard normal distribution table or a calculator to find the probability that the sample mean is greater than $11.

The correct choice from the given options for the probability that the wages for a sample of 20 receptionists exceed $11/hour is 0.0222, based on the calculated z-score and the properties of the normal distribution.

Answer 2

The probability that the wages for a sample of 20 receptionists exceeds $11/hour is approximately 0.0881.

To find the probability that the wages for a sample of 20 receptionists exceed $11/hour, we'll use the standard normal distribution.

First, let's calculate the standard error of the mean (SEM), which is the standard deviation of the sample mean distribution:

`\[\text{SEM} = \frac{\text{Standard Deviation}}{\sqrt{\text{Sample Size}}} = (2.05)/(√(20)) \approx 0.458\]`

Next, let's standardize the value $x = $11/hour using the formula:

`\[z = \frac{x - \text{Mean}}{\text{SEM}} = (11 - 10.38)/(0.458) \approx 1.356\]`

Now, we want to find the probability that the standardized value of $11/hour (represented by z = 1.356) in the standard normal distribution is greater than this value. Consulting the standard normal distribution table or using a calculator, we find the probability to be approximately 0.0881.

Therefore, the probability that the wages for a sample of 20 receptionists exceed $11/hour is approximately 0.0881. The correct choice from the given options is 0.0881.