Question

Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ = 1, (which is identical to a standard gamma distribution with α = 1), compute the following. (If necessary, round your answer to three decimal places.)
(a) The expected time between two successive arrivals
(b) The standard deviation of the time between successive arrivals
(c) P(X ? 1)
(d) P(2 ? X ? 5)

Answer 1

Answer:

(a) $E(X) = 1$

(b) $\sigma_X =1$

(c)  P(X ≤ 1) = 0.6321

(d) P(2 ≤ X ≤ 5) = 0.1286

Step-by-step explanation:

From the given information; Let consider X to be the time between two successive arrivals at the drive-up window of a local bank.

However; If X is regarded as the exponential distribution with λ = 1 which is identical to a standard gamma distribution with ∝ = 1

The objective is to compute the following :

(a) The expected time between two successive arrivals is;

$E(X) = \dfrac{1}{\lambda}$

$E(X) = \dfrac{1}{1}$

$E(X) = 1$

(b)  The standard deviation of the time between successive arrivals is;

$\sigma_X = \sqrt{\dfrac{1}{\lambda^2}}$

$\sigma_X ={\dfrac{1}{\lambda}}$

$\sigma_X ={\dfrac{1}{1}}$

$\sigma_X =1$

(c)  P(X ≤ 1)

P(X ≤ 1) = 1 - e⁻¹

P(X ≤ 1) = 0.6321

(d) P(2 ≤ X ≤ 5)

P(2 ≤ X ≤ 5) = [1 -  e⁻⁵] - [1 - e⁻²]

P(2 ≤ X ≤ 5) = 0.9933 - 0.8647

P(2 ≤ X ≤ 5) = 0.1286

Answer 2

(a) The expected time will be "1".

(b) The standard deviation will be "1".

(c) The probability be "0.6321".

(d) The probability be "0.1286".

Probability

According to the question,

Exponential distribution, λ = 1

Standard gamma distribution, α = 1

(a) Between 2 arrivals, the expected time be:

→ E(X) = $\frac{1}{\lambda}$

          = $\frac{1}{1}$

          = 1

(b) The standard deviation time be:

→ $\sigma_x$ = $\sqrt{\frac{1}{\lambda^2} }$

       = $\sqrt{\frac{1}{\lambda} }$

       = $\frac{1}{1}$

       = 1

(c) The probability be:

→ P(X $\leq$ 1) = 1 - e⁻¹

                = 0.6321

(d) The probability be:

→ P(2 $\leq$ X $\leq$ 5) = [1 - e⁻⁵] - [1 - e⁻²]

                       = 0.9933 - 0.8647

                       = 0.1286

Thus the approach above is correct.      

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