Question

Customers arrive at a particular store at the rate of 30 customers per hour.

1. What is the probability of 15 customers arriving in an hour?

a. .500
b. .001
c. .617
d. .100

2. What is the probability of 10 customers arriving in a 20-minute interval?

a. .500
b. .000
c. .125
d. 1.00

3. What is the probability of at least 2 customers arriving in a 5-minute interval?

a. .257
b. .713
c. .743
d. .287

4. What is the probability of 3 customers arriving in a 5-minute interval?

a. .000
b. .214
c. .500
d. .100

Answer 1

Answer on Question #52391 – Math – Statistics and Probability

Customers arrive at a particular store at the rate of 30 customers per hour.

The Poisson distribution with mean $m = 30$ per hour $= 2.5$ per 5 min $= 10$ per 20 min.

$$P[k] = e^{-m} \frac{m^k}{k!}$$

a. $P[3] = .2138 <---$

b. $P[\geq 3] = 1 - P[\leq 2] = .4562 <---$

c. $P[0] = .0821$

1. What is the probability of 15 customers arriving in an hour?

a. .500

b. .001

c. .617

d. .100

**Solution**

$$P[15] = e^{-30} \frac{30^{15}}{15!} = 0.001.$$

**Answer: b. .001.**

2. What is the probability of 10 customers arriving in a 20-minute interval?

a. .500

b. .000

c. .125

d. 1.00

**Solution**

$$P[10] = e^{-10} \frac{10^{10}}{10!} = 0.125.$$

**Answer: c. .125.**

3. What is the probability of at least 2 customers arriving in a 5-minute interval?

a. .257

b. .713

c. .743

d. .287

**Solution**

$$P[k \geq 2] = 1 - P[k < 2] = 1 - P[0] - P[1].$$ $$P[k \geq 2] = 1 - e^{-2.5} \frac{2.5^0}{0!} - e^{-2.5} \frac{2.5^1}{1!} = 0.713.$$

Answer: b. .713.

4. What is the probability of 3 customers arriving in a 5-minute interval?

a. .000

b. .214

c. .500

d. .100

**Solution**

$$P[3] = e^{-2.5} \frac{2.5^3}{3!} = 0.214.$$

Answer: b. .214.

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