Questions: Question 9 of 20 Step 1 of 1 01:15:31 The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean would be less than 16219 dollars if a sample of 476 persons is randomly selected? Round your answer to four decimal places.

Solution Steps

Given the variance \( \sigma^2 = 682276 \), we can find the standard deviation \( \sigma \) using the formula:

\[ \sigma = \sqrt{682276} = 826.0 \]

To find the probability that the sample mean is less than \( 16219 \), we first calculate the Z-score. The Z-score is given by:

\[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

Where:

• \( \bar{x} = 16219 \) (sample mean)
• \( \mu = 16127 \) (population mean)
• \( n = 476 \) (sample size)

Calculating the standard error:

\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{826.0}{\sqrt{476}} \approx 37.5 \]

Now, substituting the values into the Z-score formula:

\[ Z_{end} = \frac{16219 - 16127}{37.5} \approx 2.43 \]

The probability that the sample mean is less than \( 16219 \) can be expressed as:

\[ P(Z < Z_{end}) = \Phi(Z_{end}) - \Phi(-\infty) \]

Where \( \Phi(Z) \) is the cumulative distribution function of the standard normal distribution. Since \( \Phi(-\infty) = 0 \), we have:

\[ P(Z < 2.43) = \Phi(2.43) \approx 0.9925 \]

Final Answer