A software company estimates that 10% of software orders will be canceled due to unexpected technical challenges. They can deliver up to 1000 software licenses on time for their next round of orders. How many total orders can the company accept to ensure that there is at least a 95% chance that all accepted orders will be delivered on time? (10 marks)

This problem involves probability, and it can be solved using concepts related to the binomial distribution. Let's break it down step by step:

### Step 1: Define the problem variables.
- Let the total number of software orders be \( n \).
- Each order has a 10% chance of being canceled, which means that each order has a 90% chance of being successfully delivered.
- The company can only deliver up to 1000 licenses.
- We need to determine how many total orders \( n \) the company can accept to ensure at least a 95% chance that they will be able to deliver no more than 1000 licenses.

### Step 2: Use the binomial distribution model.
The number of successful deliveries \( X \) out of \( n \) orders follows a binomial distribution:

\[
X \sim \text{Binomial}(n, 0.9)
\]

Where:
- \( n \) is the number of trials (the number of orders),
- 0.9 is the probability of success (successful delivery of each order).

We want the probability that the number of successful deliveries \( X \) is less than or equal to 1000 to be at least 95%:

\[
P(X \leq 1000) \geq 0.95
\]

### Step 3: Approximate with a normal distribution.
Since \( n \) can be large, we can use the normal approximation to the binomial distribution. The mean \( \mu \) and the standard deviation \( \sigma \) of the binomial distribution are given by:

\[
\mu = n \cdot 0.9
\]
\[
\sigma = \sqrt{n \cdot 0.9 \cdot 0.1} = \sqrt{0.09n}
\]

Using the normal approximation, we want:

\[
P(X \leq 1000) \approx P\left( Z \leq \frac{1000 - \mu}{\sigma} \right) \geq 0.95
\]

From the standard normal distribution, we know that for a 95% probability, the corresponding \( Z \)-value is approximately \( Z = 1.645 \).

So, we set up the inequality:

\[
\frac{1000 - 0.9n}{\sqrt{0.09n}} \geq -1.645
\]

### Step 4: Solve the inequality.
Rearrange the inequality to solve for \( n \):

\[
\frac{1000 - 0.9n}{\sqrt{0.09n}} = -1.645
\]

Multiply both sides by \( \sqrt{0.09n} \):

\[
1000 - 0.9n = -1.645 \cdot \sqrt{0.09n}
\]

Square both sides to eliminate the square root:

\[
(1000 - 0.9n)^2 = (-1.645)^2 \cdot 0.09n
\]

\[
(1000 - 0.9n)^2 = 2.705 \cdot 0.09n
\]

\[
(1000 - 0.9n)^2 = 0.24345n
\]

Expand the left side:

\[
1000000 - 1800n + 0.81n^2 = 0.24345n
\]

Rearrange the equation into a quadratic form:

\[
0.81n^2 - 1800.24345n + 1000000 = 0
\]

### Step 5: Solve the quadratic equation.

Using the quadratic formula:

\[
n = \frac{-(-1800.24345) \pm \sqrt{(-1800.24345)^2 - 4 \cdot 0.81 \cdot 1000000}}{2 \cdot 0.81}
\]

\[
n = \frac{1800.24345 \pm \sqrt{3240875.299 - 3240000}}{1.62}
\]

\[
n = \frac{1800.24345 \pm \sqrt{875.299}}{1.62}
\]

\[
n = \frac{1800.24345 \pm 29.590}{1.62}
\]

\[
n_1 = \frac{1800.24345 + 29.590}{1.62} \approx 1131
\]
\[
n_2 = \frac{1800.24345 - 29.590}{1.62} \approx 1093
\]

Since we want a 95% chance that all orders are delivered on time, the company can accept approximately **1093** orders.

### Conclusion:
The company can accept up to **1093 orders** to ensure that there is at least a 95% chance that all accepted orders will be delivered on time.

---

Do you want more details on any step or clarification? 

Here are some related questions:
1. How does the normal approximation work for binomial distributions?
2. What would happen if the probability of order cancellation increased?
3. How can the result be verified without using the normal approximation?
4. What does the 95% confidence level mean in practical terms for this problem?
5. How would the calculation change if the company could only deliver 900 licenses?

**Tip:** In probability problems, using normal approximations for binomial distributions simplifies computations when \( n \) is large.

A software company estimates that 10% of software orders will be canceled due to unexpected technical challenges. They can deliver up to 1000 software licenses on time for their next round of orders. How many total orders can the company accept to ensure that there is at least a 95% chance that all accepted orders will be delivered on time?

Solve a probability problem involving a software company estimating a 10% cancellation rate for orders. Learn how to use the binomial distribution and normal approximation to determine the maximum number of orders that ensure a 95% chance of delivering no more than 1000 licenses. This solution includes step-by-step calculations and the application of the quadratic formula.

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