A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each one recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order. Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

Answer:Step-by-step explanation:Hello!X: number of absences per tutorial per student over the past 5 years(percentage)X≈N(μ;σ²)You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.The formula for the CI is:X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4X[bar]= 10.41S= 3.71[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.645[/tex][10.41±1.645*[tex](\frac{3.71}{\sqrt{28} } )[/tex]][9.26; 11.56]Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.I hope this helps!