Step 1
A) The mean distribution
[tex]\frac{1+53}{2}=\frac{54}{2}=27.0000[/tex]Step 2
B) The standard deviation
[tex]\begin{gathered} SD=\sqrt[]{\frac{1}{12}\times(b-a)^2} \\ SD=\sqrt[]{\frac{1}{12}(53-1)^2} \\ SD=\text{ }15.0111 \end{gathered}[/tex]Step 3
C)
[tex]P(x=18)=0[/tex]Step 4
D)
[tex]\begin{gathered} P(10Step 5E)
[tex]P(x>35)=\text{ }\frac{53-35}{52}=\frac{18}{52}=0.3462[/tex]Step 6
F)
[tex]P(x>18|x<32)=\text{ }\frac{32-18}{32-1}=\frac{14}{31}=0.4516[/tex]Step 7
G)
[tex]\begin{gathered} \text{The 47th percentile}=1\text{ + }\frac{47}{100}(53-1)_{} \\ =1+0.47(52)=25.44_{}00 \end{gathered}[/tex]Step 8
[tex]\begin{gathered} \text{The minimum for the upper percentile = 1+((}\frac{3}{4})(53^{}-1) \\ =1+0.75(52) \\ =1+\text{ 39=40}.0000 \end{gathered}[/tex]