Respuesta :
Answer:
Option A - The population mean is [tex]\mu\approx 81.36[/tex]
Confidence interval is (3.153,166.047).
Step-by-step explanation:
Given : Assume that the population has a normal distribution
n=30, x=84.6, s=10.5, 90% confidence.
To find : Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu?
Solution :
First we apply z-score formula,
[tex]z=\frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
Where, x is the sample mean x=84.6
s is the standard deviation s=10.5
n is the number of sample n=30
[tex]\mu[/tex] is the population mean
z is the score value, at 90% z=1.645
Substitute all the values in the formula,
[tex]1.645=\frac{84.6-\mu}{\frac{10.5}{\sqrt{30}}}[/tex]
[tex]1.645\times \frac{10.5}{\sqrt{30}}=84.6-\mu[/tex]
[tex]1.645\times 1.917=84.6-\mu[/tex]
[tex]3.153=84.6-\mu[/tex]
[tex]\mu=84.6-3.153[/tex]
[tex]\mu=81.447[/tex]
So, The population mean is [tex]\mu=81.447[/tex]
Therefore, Option A is correct.
Now, Apply confidence interval formula
[tex]x-\mu<CI<x+\mu[/tex]
[tex]84.6-81.447<CI<84.6+81.447[/tex]
[tex]3.153<CI<84.6+166.047[/tex]
Therefore, Confidence interval is (3.153,166.047).
