Answer
$P(fewer~than~seven)=P(x\lt7)=0.450$
On about 450 of every 1000 days, there will be fewer than 7 hits to a Web site between 7:30 p.m. and 7:35 p.m.
Work Step by Step
$P(fewer~than~seven)=P(x\lt7)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=\frac{(1.4\times5)^{0}}{0!}e^{-1.4\times5}+\frac{(1.4\times5)^{1}}{1!}e^{-1.4\times5}+\frac{(1.4\times5)^{2}}{2!}e^{-1.4\times5}+\frac{(1.4\times5)^{3}}{3!}e^{-1.4\times5}+\frac{(1.4\times5)^{4}}{4!}e^{-1.4\times5}+\frac{(1.4\times5)^{5}}{5!}e^{-1.4\times5}+\frac{(1.4\times5)^{6}}{6!}e^{-1.4\times5}=\frac{1}{1}e^{-7}+\frac{7}{1}e^{-7}+\frac{49}{2}e^{-7}+\frac{343}{6}e^{-7}+\frac{2401}{24}e^{-7}+\frac{16807}{120}e^{-7}+\frac{117649}{720}e^{-7}=0.450$
