Answer: a. 19.200 to 20.800
Step-by-step explanation:
Formula to find the confidence interval[tex](\mu)[/tex] :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n is the sample size
[tex]\sigma[/tex] = Population standard deviation.
[tex]\overline{x}[/tex]= Sample mean
[tex]z_{\alpha/2}[/tex] = Two tailed z-value for significance level of [tex]\alpha[/tex].
Given : Confidence level = 95.44% = 0.9544
Significance level = [tex]\alpha=1-0.9544=0.0456[/tex]
Using standard normal z-value table ,
Two tailed z-value for Significance level of 0.0456 :
[tex]z_{\alpha/2}=z_{0.0228}=1.999\approx2[/tex]
Also,
n=144
[tex]\sigma= 4.8[/tex]
[tex]\overline{x}=20[/tex]
Then, the required 95.44% confidence interval for the population mean :-
[tex]20\pm (2)\dfrac{4.8}{\sqrt{144}}\\\\ =20\pm (0.800)\\\\=(20-0.800,\ 20+0.800)=(19.200,\ 20.800)[/tex]
Hence, the 95.44% confidence interval for the population mean is 19.200 to 20.800.
