A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8. The 95.44% confidence interval for the population mean is______? a. 19.200 to 20.800 b. 18.78 to 20.784 c. 15.2 to 24.8 d. 19.216 to 20.784 e. 21.2 to 22.8

Answer: a. 19.200 to 20.800Step-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% = 0.9544Significance 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.