Question 1178059
For the Poisson distribution, {{{mu = 100}}}, and {{{sigma^2 = 100}}}, and so {{{sigma = 10}}}.

Chebyshev's theorem says that {{{P(abs(X - mu)<= k*sigma) >= 1-1/k^2}}}.

==> {{{P(abs(X - 100) <= 10k) >= 1-1/k^2}}}, and k has to be determined.

{{{abs(X - 100) <= 10k }}} is the same as {{{-10k <= X - 100 <= 10k}}}, or {{{70 = 100-10k <= X <= 100+10k = 130}}}.

This implies that k = 3.    We then get {{{P(abs(X - 100) <= 30) >= 1-1/3^2 = 8/9}}}.

Therefore a lower bound for probability of cars arriving at the intersection in 1 hour is between 70 and 130 is {{{highlight(8/9)}}}.