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 [tex] \$98.75 [/tex] with a standard deviation of [tex] \$10.45 [/tex]. Susan has a sample size of 60.

Complete the equation that Susan can use to find the interval in which she can be [tex] 68\% [/tex] sure that the sample mean will lie.

[tex]\[
\begin{array}{llllll}
7.75 & 98.75 & 88.3 & 10.45 & 109.2 & 60
\end{array}
\][/tex]

Drag the values to the correct locations on the image. Not all values will be used.



Answer :

Susan is conducting a survey to understand the electric bills of households in her city. She finds that the average (mean) electric bill for the population is [tex]$98.75$[/tex] with a standard deviation of [tex]$10.45$[/tex]. With a sample size of [tex]$60$[/tex] households, Susan wants to find the interval in which she can be [tex]$68\%$[/tex] confident that the sample mean will lie.

To do this, Susan needs to calculate the confidence interval for the sample mean. The steps involved in calculating this are as follows:

1. Find the z-score corresponding to the confidence level: For a [tex]$68\%$[/tex] confidence level, the z-score is approximately [tex]$1$[/tex]. This is because [tex]$68\%$[/tex] is within [tex]$1$[/tex] standard deviation of the mean in a normal distribution.

2. Calculate the Standard Error (SE) of the sample mean: The formula for the standard error is
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]
where [tex]$\sigma$[/tex] is the population standard deviation and [tex]$n$[/tex] is the sample size.
Here,
[tex]\[ \text{SE} = \frac{10.45}{\sqrt{60}} \approx 1.3491 \][/tex]

3. Calculate the Margin of Error (MOE): The margin of error is computed using the z-score and the standard error:
[tex]\[ \text{MOE} = z \times \text{SE} \][/tex]
With [tex]$z = 1$[/tex] and [tex]$\text{SE} \approx 1.3491$[/tex],
[tex]\[ \text{MOE} = 1 \times 1.3491 \approx 1.3491 \][/tex]

4. Calculate the Confidence Interval: The confidence interval is given by
[tex]\[ (\text{Population mean} - \text{MOE}, \text{Population mean} + \text{MOE}) \][/tex]
Thus,
[tex]\[ (98.75 - 1.3491, 98.75 + 1.3491) = (97.4009, 100.0991) \][/tex]

Therefore, to complete the equation that Susan can use to find the interval in which she can be [tex]$68\%$[/tex] sure that the sample mean will lie, we can fill in the following values:

- Population mean ([tex]$\bar{x}$[/tex]): [tex]$98.75$[/tex]
- Standard deviation ([tex]$\sigma$[/tex]): [tex]$10.45$[/tex]
- Sample size ([tex]$n$[/tex]): [tex]$60$[/tex]

So the equation for the [tex]$68\%$[/tex] confidence interval is:
[tex]\[ (98.75 - 1.3491, 98.75 + 1.3491) = (97.4009, 100.0991) \][/tex]