Answer
$P(fewer~than~eight)=P(x\lt8)=0.179$
On about 179 of every 1000 days, there will be fewer than 8 calls between 3:10 p.m. and 3:15 p.m.
Work Step by Step
$P(fewer~than~eight)=P(x\lt8)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)=\frac{(2.1\times5)^{0}}{0!}e^{-2.1\times5}+\frac{(2.1\times5)^{1}}{1!}e^{-2.1\times5}+\frac{(2.1\times5)^{2}}{2!}e^{-2.1\times5}+\frac{(2.1\times5)^{3}}{3!}e^{-2.1\times5}+\frac{(2.1\times5)^{4}}{4!}e^{-2.1\times5}+\frac{(2.1\times5)^{5}}{5!}e^{-2.1\times5}+\frac{(2.1\times5)^{6}}{6!}e^{-2.1\times5}+\frac{(2.1\times5)^{7}}{7!}e^{-2.1\times5}=\frac{1}{1}e^{-10.5}+\frac{10.5}{1}e^{-10.5}+\frac{110.25}{2}e^{-10.5}+\frac{1157.625}{6}e^{-10.5}+\frac{12155.0625}{24}e^{-10.5}+\frac{127628.15625}{120}e^{-10.5}+\frac{1340095.640625}{720}e^{-10.5}+\frac{14071004.2265625}{5040}e^{-10.5}=0.179$
