Answer to Question #182837 in Statistics and Probability for Junied Mohammed Tazrian

Question

In section A the marks of quiz 1 of 6 students are randomly picked as 12,8,17,9,11,1 what is left limit of mean at 95% confidence level

Answer 1

$M = \frac{12+8+17 +9+11+1}{6} = 9.67$

$SD = \sqrt{\dfrac{\sum{\vert x-M\vert}^2}{N}}=\sqrt{\dfrac{(12-9.67)^2+(8-9.67)^2+...+(1-9.67)^2}{6}}=4.82$

95% confidence level means that Z = 1.96

$s_M= \sqrt{4.82^2/6}=1.97$

μ = M ± Z(s_M)

μ = 9.67 ± 1.96*1.97

μ = 9.67 ± 3.86

Left limit: 9.67 - 3.86 = 5.81