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At a $911$call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has an exponential distribution.

From the given information, the average rate of one call is for every two minutes. Therefore, time occurs between five consecutive calls is given by$2脳5=10mins$

The cumulative distribution function is $P(T<t)=1鈭�e^{鈭�\frac{t}{2}}$

Therefore, $P(T>3)=1-P(T<3)=1-(1鈭�e^{鈭�\frac{3}{2}})=0.2231$

the probability that after a call is received, it takes more than three minutes for the next call to occur. is$0.2231$

We want to find $0.9=P(T>t)=1鈭�\left(1鈭�e^{鈭�\frac{t}{2}}\right)=e^{鈭�\frac{t}{2}}$

Solving for, ${\mathrm{e}}^{鈭�\frac{t}{2}}=0.90,\text{so}鈭�\frac{t}{2}=\ln 鈦�\left(0.90\right),\text{and}t=鈭�2\ln 鈦�\left(0.90\right)鈮�\text{0.21mins.}$

Ninety percent of all calls occur within$0.21$ minutes of the previous call

We want to find $P(1<t<2).$

To do this, $P(1<t<2)鈥�P(t<1)$

$=\left(1鈭�e^{鈭�\frac{1}{2}鈭�2}\right)鈭�\left(1鈭�e^{鈭�\frac{1}{2}鈭�1}\right)鈮�0.6065鈭�0.3678=0.2387$

the probability that the next call will occur within the next minute is$0.238$ mins

number of calls $n=20$

we know the cdf, $P(T<t)=1鈭�e^{鈭�\frac{t}{2}}$

The probability of a call occur within an hour is given by,

$P(T<60)=1鈭�e^{鈭�\frac{60}{2}}=1$

The probability of$20$calls occur within an hour is given by$nP(1-P)=20脳1脳0=0$