Ten percent of the population is left-handed. Approximate the probability that the number of students in INDENG 241 who are left-handed is between 21 and 23 (inclusive). There are currently 219 students enrolled in the course. State your assumptions.

To approximate the probability that the number of left-handed students in INDENG 241 is between 21 and 23 (inclusive), we can model this scenario using the **binomial distribution**. Here's a step-by-step solution:

### Step 1: Assumptions
1. **Probability of being left-handed (p):** This is given as 10% (or 0.10).
2. **Total number of students (n):** There are 219 students enrolled in INDENG 241.
3. **Independence:** We assume that whether one student is left-handed is independent of whether another student is left-handed.

### Step 2: Define the Binomial Distribution
The number of left-handed students \(X\) follows a binomial distribution:
\[
X \sim \text{Binomial}(n = 219, p = 0.10)
\]
This means the probability mass function (PMF) is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where \(n\) is the total number of students, \(p\) is the probability of being left-handed, and \(k\) is the number of left-handed students.

### Step 3: Normal Approximation to the Binomial
Since \(n = 219\) is large and \(p = 0.10\) is neither too small nor too large, we can approximate the binomial distribution using the **normal distribution** with the following parameters:
- Mean \( \mu = np = 219 \times 0.10 = 21.9 \)
- Standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{219 \times 0.10 \times 0.90} \approx 4.44 \)

Thus, \( X \) is approximately distributed as:
\[
X \sim N(\mu = 21.9, \sigma = 4.44)
\]

### Step 4: Apply the Continuity Correction
To find the probability that the number of left-handed students is between 21 and 23 (inclusive), we apply the continuity correction for discrete values. We will compute the probability for the range \( 20.5 \leq X \leq 23.5 \) using the normal distribution.

### Step 5: Standardize the Normal Distribution
We now standardize the values to the standard normal distribution \(Z \sim N(0, 1)\):
- For \(X = 20.5\):
\[
Z_1 = \frac{20.5 - 21.9}{4.44} \approx -0.31
\]
- For \(X = 23.5\):
\[
Z_2 = \frac{23.5 - 21.9}{4.44} \approx 0.36
\]

### Step 6: Use the Standard Normal Table
Using the standard normal distribution table (or a calculator), we find:
- \( P(Z \leq -0.31) \approx 0.3783 \)
- \( P(Z \leq 0.36) \approx 0.6406 \)

Thus, the probability that the number of left-handed students is between 21 and 23 is:
\[
P(20.5 \leq X \leq 23.5) = P(Z \leq 0.36) - P(Z \leq -0.31) \approx 0.6406 - 0.3783 = 0.2623
\]

### Final Answer:
The approximate probability that the number of left-handed students in INDENG 241 is between 21 and 23 (inclusive) is **0.2623** or **26.23%**.

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Do you want more details or have any questions?

Here are some related questions to further expand this concept:
1. How would the result change if the probability of being left-handed was higher, say 15%?
2. What is the probability that fewer than 20 students are left-handed?
3. How would you compute the probability using the exact binomial formula rather than the normal approximation?
4. What happens to the normal approximation as the sample size decreases?
5. How does the continuity correction affect the accuracy of the normal approximation?

**Tip:** The continuity correction is important when using the normal approximation for a discrete distribution because it adjusts for the fact that the normal distribution is continuous, while the binomial distribution is not.

Approximate the probability that the number of left-handed students in a class of 219 is between 21 and 23 using the binomial distribution and normal approximation. This solution covers key probability concepts, including continuity correction, and is suitable for undergraduate students.

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