Questions: Susan is conducting a survey about the electric bills of households in her city. From the electric company, she found that the population mean is 98.75 with a standard deviation of 10.45. Susan has a sample size of 60. Complete the equation that Susan can use to find the interval in which she can be 68% sure that the sample mean will lie. 7.75 60 10.45 109.2 88.3 98.75

Susan is conducting a survey about the electric bills of households in her city. From the electric company, she found that the population mean is 98.75 with a standard deviation of 10.45. Susan has a sample size of 60.

Complete the equation that Susan can use to find the interval in which she can be 68% sure that the sample mean will lie.
7.75 60 10.45 109.2 88.3 98.75

Transcript text: Susan is conducting a survey about the electric bills of households in her city. From the electric company, she found that the population mean is $\$ 98.75$ with a standard deviation of $\$ 10.45$. Susan has a sample size of 60. Complete the equation that Susan can use to find the interval in which she can be $68 \%$ sure that the sample mean will lie. \[ \begin{array}{llllll} 7.75 & 60 & 10.45 & 109.2 & 88.3 & 98.75 \end{array} \]

Solution

Solution Steps

Step 1: Find the standard error

The standard error is calculated by dividing the standard deviation by the square root of the sample size. In this case, the standard deviation is $10.45 and the sample size is 60. So, the standard error is $\frac{10.45}{\sqrt{60}} \approx 1.35$.

Step 2: Calculate the interval

For a 68% confidence interval, we use one standard error above and below the population mean. So the interval is $98.75 \pm 1.35$.