A simple random sample is drawn from a normally distributed population. The margin of error is 5.9 at a 95% level of confidence. If the sample mean is 18.7, what is the 95% confidence interval for the population mean?

A. [tex]\( 18.7 \pm 5.9 \)[/tex]
B. [tex]\( 18.7 \pm 9.7 \)[/tex]
C. [tex]\( 18.7 \pm 11.6 \)[/tex]
D. [tex]\( 18.7 \pm 15.2 \)[/tex]



Answer :

To determine the [tex]$95\%$[/tex] confidence interval for the population mean based on the provided information, follow these steps:

1. Identify the given values:
- The sample mean ([tex]\(\bar{x}\)[/tex]) is 18.7.
- The margin of error (E) is 5.9.

2. Recall the formula for the confidence interval:
The confidence interval [tex]\( CI \)[/tex] for a population mean is given by:
[tex]\[ CI = \left( \bar{x} - E, \bar{x} + E \right) \][/tex]

3. Substitute the given values into the formula:
[tex]\[ CI = \left( 18.7 - 5.9, 18.7 + 5.9 \right) \][/tex]

4. Perform the calculations:
- Calculate the lower bound:
[tex]\[ 18.7 - 5.9 = 12.8 \][/tex]
- Calculate the upper bound:
[tex]\[ 18.7 + 5.9 = 24.6 \][/tex]

5. Combine the results:
Thus, the [tex]$95\%$[/tex] confidence interval for the population mean is:
[tex]\[ (12.8, 24.6) \][/tex]

Given the options provided:
- [tex]\( 18.7 \pm 5.9 \)[/tex]
- [tex]\( 18.7 \pm 9.7 \)[/tex]
- [tex]\( 18.7 \pm 11.6 \)[/tex]
- [tex]\( 18.7 \pm 15.2 \)[/tex]

It is clear that the correct option is [tex]\( 18.7 \pm 5.9 \)[/tex].

The [tex]$95\%$[/tex] confidence interval for the population mean is [tex]\( 18.7 \pm 5.9, \)[/tex] which corresponds to the interval [tex]\( (12.8, 24.6) \)[/tex].