In a certain store, there is a .03 probability that the scanned price in the bar code scanner will not match the advertised price. The cashier scans 780 items. What is the expected number of mismatches? What is the standard deviation rounded to four decimal places? What is the approximate normal probability of atleast 18 mismatches rounded to four decimal places? What is the probability for more than 32?

To solve this problem, we need to understand the principles of probability and statistics. Let's break down each part step-by-step.

### Expected Number of Mismatches

The expected number of mismatches can be calculated using the formula for the expected value of a binomial distribution:

\[ E(X) = n \times p \]

Where:
- \( n \) is the number of trials (780 items)
- \( p \) is the probability of a mismatch (0.03)

So,

\[ E(X) = 780 \times 0.03 = 23.4 \]

### Standard Deviation

The standard deviation for a binomial distribution is given by:

\[ \sigma = \sqrt{n \times p \times (1 - p)} \]

So,

\[ \sigma = \sqrt{780 \times 0.03 \times (1 - 0.03)} = \sqrt{780 \times 0.03 \times 0.97} = \sqrt{22.698} \approx 4.7637 \]

### Normal Approximation

To find the approximate normal probability of at least 18 mismatches, we use the normal approximation to the binomial distribution. For this, we need the z-score:

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

First, let's calculate the z-score for \( X = 18 \):

\[ \mu = 23.4 \]
\[ \sigma = 4.7637 \]

\[ z = \frac{18 - 23.4}{4.7637} = \frac{-5.4}{4.7637} \approx -1.1333 \]

Using the z-table, the probability corresponding to \( z = -1.1333 \) is approximately 0.1292. Therefore, the probability of having less than 18 mismatches is 0.1292. Thus, the probability of having at least 18 mismatches is:

\[ P(X \geq 18) = 1 - P(X < 18) = 1 - 0.1292 = 0.8708 \]

### Probability of More Than 32 Mismatches

For more than 32 mismatches:

First, let's calculate the z-score for \( X = 32 \):

\[ z = \frac{32 - 23.4}{4.7637} = \frac{8.6}{4.7637} \approx 1.8051 \]

Using the z-table, the probability corresponding to \( z = 1.8051 \) is approximately 0.9646. Therefore, the probability of having 32 or fewer mismatches is 0.9646. Thus, the probability of having more than 32 mismatches is:

\[ P(X > 32) = 1 - P(X \leq 32) = 1 - 0.9646 = 0.0354 \]

### Summary

- The expected number of mismatches is \( 23.4 \).
- The standard deviation is \( 4.7637 \) (rounded to four decimal places).
- The approximate normal probability of at least 18 mismatches is \( 0.8708 \) (rounded to four decimal places).
- The probability of more than 32 mismatches is \( 0.0354 \) (rounded to four decimal places).

Would you like more details or have any questions about this problem?

Here are some additional related questions for further exploration:

1. What is the probability of exactly 23 mismatches?
2. How would the expected number of mismatches change if the number of items scanned increased to 1000?
3. If the probability of a mismatch increased to 0.05, what would be the new expected number of mismatches?
4. How would the standard deviation change if the probability of a mismatch decreased to 0.01?
5. What is the probability of having between 20 and 25 mismatches inclusive?
6. How does the shape of the binomial distribution change as the number of items scanned increases?
7. What is the probability of having fewer than 10 mismatches?
8. If the store implements a new scanning system that reduces the mismatch probability to 0.01, what is the probability of having more than 15 mismatches?

**Tip:** When dealing with large sample sizes and small probabilities, the normal approximation to the binomial distribution can be a very useful tool for simplifying complex probability calculations.

Learn how to calculate the expected number of mismatches and probability distributions in a scenario involving probability and statistics. Solve for the expected mismatches, standard deviation, and probabilities using binomial and normal distributions. Suitable for advanced high school students.

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