Question

drag the values to the correct locations on the image. not all values will be used. 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 98.75 88.3 10.45 109.2

Explanation:

Step1: Recall the formula for confidence - interval of sample mean

For a normal distribution, when the population standard - deviation $\sigma$ is known, the formula for the confidence interval of the sample mean is $\bar{x}\pm z\frac{\sigma}{\sqrt{n}}$. In the case of a 68% confidence interval, the $z$ - value is approximately 1 (since about 68% of the data in a normal distribution lies within 1 standard deviation of the mean). Here, we are interested in the form of the formula with the population mean $\mu$ as the center, so it is $\mu\pm\frac{\sigma}{\sqrt{n}}$.

Step2: Identify the values of $\mu$, $\sigma$, and $n$

The population mean $\mu = 98.75$, the population standard deviation $\sigma=10.45$, and the sample size $n = 60$.

Step3: Fill in the formula

The formula for the interval is $98.75\pm\frac{10.45}{\sqrt{60}}$.

Answer:

The left - hand box: 98.75; the top box in the fraction: 10.45; the bottom box in the square - root: 60.