Answer:
[tex]99.3<\mu<104.9[/tex]
Therefore, option D is correct.
Step-by-step explanation:
We have been given [tex]n=62,\sim=11.4\text{and}\bar{x}=102.1[/tex]
The formula to find interval is:[tex]\mu[/tex] which is unknown mean.
[tex]\bar{x}\pm\frac{\sim}{\sqrt{n}}\cdot \text{z-score}[/tex]
So, 95% confidence interval is standard value of z-score at 95% confidence interval is 1.96
Substituting the values in the formula we will get:
[tex]102.1/pm\frac{11.4}{\sqrt{62}}(1.96)[/tex]
On simplifying the above equation we will get:
Taking [tex]102.1-\frac{11.4}{\sqrt{62}}(1.96)=99.26≈99.3[/tex]
and [tex]102.1+\frac{11.4}{\sqrt{62}}(1.96)=104.93[/tex]
Therefore, option D is correct.
[tex]99.3<\mu<104.9[/tex]
