Question 3.15: Refer to Figure 3.1 (B(12, 1/4)) and determine x0.25, x0.50,......
Refer to Figure 3.1 (B(12, 1/4)) and determine x_{0.25},x_{0.50},\mathrm{and}\;x_{0.75}.
Here \,x_{0.25}=2 since P(X\lt 2)=P(X=0)+P(X=1)= 0.1584\leq0.25~{\mathrm{and}}\;P(X\leq2)=0.1584+P(X=2)=0.3907\geq0.25. Likewise, x_{0.50}={{{3}}}~\mathrm{sinc e}\;P(X\lt {{{3}}}){=}0{.}3907\leq0.50\mathrm{~and~}P(X\leq3){=}0.6488\ge0.50. Finally, x_{0.75}=4,\mathrm{since}\;P(X\lt 4)=0.6488\leq0.75\;\mathrm{and}\;P(X\leq4)=0.8424\gt 0.75.
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