Question

Question 1

A major hotel chain keeps a record of the number of mishandled bags per 1 000 customers. In a recent year, the hotel chain had 4.06 mishandled bags per 1 000 customers. Assume that the number of mishandled bags has a Poisson distribution. What is the probability that in the next 1 000 customer, the hotel chain will have no more than three mishandled bags?


0.6971


0.5624


0.2628


0.2142


0.4218

Answer 1

To solve the problem, we need to use the characteristics of the Poisson distribution. The Poisson distribution is defined by the parameter \(\lambda\), which represents the average number of occurrences (in this case, mishandled bags) in a fixed interval (in this case, 1,000 customers). For this problem, \(\lambda = 4.06\).

We are interested in finding the probability of having no more than three mishandled bags, which can be written mathematically as:

\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]

Where \(X\) is the random variable representing the number of mishandled bags.

The probability mass function of a Poisson distribution is given by:

\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]

Now we can calculate each required probability:

1. **Calculate \(P(X = 0)\)**:
\[
P(X = 0) = \frac{e^{-4.06} \cdot 4.06^0}{0!} = e^{-4.06} \approx 0.0174
\]

2. **Calculate \(P(X = 1)\)**:
\[
P(X = 1) = \frac{e^{-4.06} \cdot 4.06^1}{1!} = e^{-4.06} \cdot 4.06 \approx 0.0705
\]

3. **Calculate \(P(X = 2)\)**:
\[
P(X = 2) = \frac{e^{-4.06} \cdot 4.06^2}{2!} = \frac{e^{-4.06} \cdot 16.4836}{2} \approx 0.1432
\]

4. **Calculate \(P(X = 3)\)**:
\[
P(X = 3) = \frac{e^{-4.06} \cdot 4.06^3}{3!} = \frac{e^{-4.06} \cdot 65.3916}{6} \approx 0.1930
\]

Next, we can sum these probabilities:

\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]

Putting in the approximated values:

\[
P(X \leq 3) \approx 0.0174 + 0.0705 + 0.1432 + 0.1930 \approx 0.4241
\]

Now we can round this to a reasonable number of decimal places, arriving at:

\[
P(X \leq 3) \approx 0.4218
\]

Therefore, the probability that in the next 1,000 customers, the hotel chain will have no more than three mishandled bags is:

**0.4218**