Question

1. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2) Find P(x > 4)

a. 0.225

b. 0.449

c. 0.551

d. 0.635

2. At an urgent care facility, patients arrive at an average rate of one patient every seven minutes.

Assume that the duration between arrivals is exponentially distributed.

Find the probability that the time between two successive visits to the urgent care facility is between three and eight minutes.

a. 0.117

b. 0.467

c. 0.667

d. 0.333

Answer 1

1.

The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2). We know that CDF for exponential distribution is given by,

$P(X\le x) = 1-e^{-\lambda x}$

$P(X>4 ) = 1-P(X \le 4)$

  

  

  

=1-1+e-0.8 0.4493 = 0.449

Ans.:- b. 0.449

2.

At an urgent care facility, patients arrive at an average rate of one patient every seven minutes.Assume that the duration between arrivals is exponentially distributed.

Here required conditions follow exponential with  $\frac{1}{7}$

X~ Exp (1/7)

Probability that the time between two successive visits to the urgent care facility is between three and eight minutes,

P( 3 < X < 8 ) = P(X < 8) - P(X < 3)

P (X< 8) = $1-e^{-\lambda X} = 1-e^{(-1/7)*8} =$ 0.681083

P(X<3) = $1-e^{-\lambda X} = 1-e^{(-1/7)*3} =$ 0.348560

P( 3 < X < 8 ) = 0.681083 - 0.348560 = 0.332523 = 0.333

Ans.:- d. 0.333