The number of bags lost by a small airline per week is approximately normally distributed with a mean of 427 bags and a standard deviation of 35 bags. What is the probability that they

a) lose exactly 430 bags next week?

b) lose at most 420 bags next week?

c) lose more than 450 bags next week?

d) lose between 440 bags and 480 bags next week?

We can solve these probability questions using the properties of the normal distribution. First, let's outline the given information:

- Mean (μ) = 427 bags
- Standard deviation (σ) = 35 bags

The standard normal distribution \( Z \) can be used by converting a raw score \( X \) into a standard score \( Z \) using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

Let's now go step by step for each part:

---

### a) Probability of losing exactly 430 bags next week?

For a continuous probability distribution (like the normal distribution), the probability of any single exact value is technically 0. This is because there are infinitely many possible values in a continuous range. Hence:

\[
P(X = 430) = 0
\]

---

### b) Probability of losing **at most 420 bags** next week?

We need to find:

\[
P(X \leq 420)
\]

First, calculate the corresponding Z-score for 420:

\[
Z = \frac{420 - 427}{35} = \frac{-7}{35} = -0.2
\]

Now, using a standard normal distribution table or a calculator, we find the probability corresponding to \( Z = -0.2 \):

\[
P(Z \leq -0.2) \approx 0.4207
\]

Thus:

\[
P(X \leq 420) \approx 0.4207
\]

This means there is about a 42.07% chance of losing at most 420 bags next week.

---

### c) Probability of losing **more than 450 bags** next week?

We need to find:

\[
P(X > 450)
\]

First, calculate the corresponding Z-score for 450:

\[
Z = \frac{450 - 427}{35} = \frac{23}{35} \approx 0.6571
\]

Now, using a standard normal distribution table or a calculator, the probability for \( Z \leq 0.6571 \) is approximately:

\[
P(Z \leq 0.6571) \approx 0.7441
\]

Thus, the probability of losing more than 450 bags is:

\[
P(X > 450) = 1 - P(Z \leq 0.6571) = 1 - 0.7441 = 0.2559
\]

This means there is about a 25.59% chance of losing more than 450 bags next week.

---

### d) Probability of losing **between 440 and 480 bags** next week?

We need to find:

\[
P(440 \leq X \leq 480)
\]

First, calculate the Z-scores for 440 and 480:

For 440:

\[
Z_1 = \frac{440 - 427}{35} = \frac{13}{35} \approx 0.3714
\]

For 480:

\[
Z_2 = \frac{480 - 427}{35} = \frac{53}{35} \approx 1.5143
\]

Now, using a standard normal distribution table or a calculator, the probabilities are:

\[
P(Z \leq 0.3714) \approx 0.6443
\]
\[
P(Z \leq 1.5143) \approx 0.9345
\]

Therefore, the probability of losing between 440 and 480 bags is:

\[
P(440 \leq X \leq 480) = P(Z \leq 1.5143) - P(Z \leq 0.3714)
\]
\[
P(440 \leq X \leq 480) \approx 0.9345 - 0.6443 = 0.2902
\]

Thus, there is about a 29.02% chance of losing between 440 and 480 bags next week.

---

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. What is the probability of losing fewer than 410 bags next week?
2. What is the probability of losing more than 470 bags next week?
3. Find the probability of losing between 430 and 450 bags.
4. What is the Z-score for losing exactly 435 bags?
5. How many bags would the airline lose with a probability of 95%?

**Tip:** Remember, in continuous distributions, the probability of getting an exact value is always 0!

The number of bags lost by a small airline per week is approximately normally distributed with a mean of 427 bags and a standard deviation of 35 bags. What is the probability that they: a) lose exactly 430 bags next week? b) lose at most 420 bags next week? c) lose more than 450 bags next week? d) lose between 440 bags and 480 bags next week?

Calculate probabilities for the number of bags lost per week by a small airline using the normal distribution. This example, suitable for advanced high school or college-level students, explores finding probabilities of losing exactly 430 bags, at most 420 bags, more than 450 bags, and between 440 and 480 bags with a mean of 427 bags and a standard deviation of 35 bags.

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