(a) What is the distribution for \(T_r\) ?

Let total number of minutes(trials) be \(n\), with \(r\) minutes as successes and \((n-r)\) minutes were no customer arrives.

Probability of customer arriving(success) is \(p\) and probability of failure is \(q = 1-p\).

Probability of no customer arriving in a unit time is \(q\) and customer arrival \(p = 1 -q\). Total number of minutes were no customer arrives = \(n-r\).

Using Bernoulli Random Variable method, \(P(T_r)\) = \(q^{n-r}p^{r}\)

In a given time, success can happen in \(n-r\) different ways = \(n\choose {n-r}\)

Therefore distribution of customer not arriving in \(r\) minutes is \(T_r\)= \(b(n, p, {n-r}) = {n\choose {n-r}}q^{n-r}p^{r}\).

\(T_r\)= \(b(n, p, {n-r}) = {n\choose {n-r}}p^{r}q^{n-r}\).

(b) What is the distribution for \(C_r\) ?

Probability of customer arrival in a unit time is success \(p\) and no customer arrival is failure \(q = 1-p\). Total number of minutes in which customer arrived = \(r\).

Using Bernoulli Random Variable method, \(P(C_r)\) = \(p^{r}q^{n-r}\)

In a given time, success can happen in \(r\) different ways = \(n\choose r\)

Therefore distribution of customer arriving in \(r\) minutes \(C_r\)= \(b(n, p, r) = {n\choose r}p^{r}q^{n-r}\).

(c) Find the mean and variance for the number of customers arriving in the first r minutes.

According Bernoulli distribution, \(\mu = E(X) = p\)

It means customer may arrive in a given minute taking the value of outcome \(1\) and \(0\) as no arrvial.

Probability of \(P(X=0) = q = 1 -p\) and probability of \(P(X=1) = p\)

\(E(X) = P(X=0)*0 + P(X=1)*1\)

\(E(X) = P(X=1)\)

\(E(X) = p\)

Hence \(\mu = p\)

Mean for the number of customers arriving in \(r\) minutes \(\mu_r = rp\).

Variance of Bernoulli distribution, \(\sigma^2 = V(X) = p(1-p)\)

\(V(X)\) is average squared distince from mean.

\(V(X) = P(X=0)*(0-\mu)^2 + P(X=1)*(1-\mu)^2\)

\(V(X) = (1-p)*(0-\mu)^2 + p*(1-\mu)^2\)

Since \(\mu = p\), \(V(X) = (1-p)*(0-p)^2 + p*(1-p)^2\)

\(V(X) = (1-p)*p^2 + p*(1-p)^2\)

\(V(X) = p^2-p^3 + p*(1-2p+p^2)\)

\(V(X) = p^2-p^3 + p-2p^2+p^3\)

\(V(X) = p(1-p)\)

Further \(1-p = q\), \(V(X) = pq\)

Variance for the number of customers arriving in the first \(r\) minutes \(\sigma^2_r = rpq\).

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