The weekly demand for General Motors (GM) car sales follows a normal distribution with a mean of 40,000 cars and a standard deviation of 12,000 cars.

(a) What is the probability that a randomly selected weekly demand for GM is less than 46,000?

To solve this problem, we need to find the probability that the weekly demand is less than 46,000. Since the demand follows a normal distribution, we will use the Z-score formula:

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

where:
- \( X \) is the value we are interested in (46,000 cars),
- \( \mu \) is the mean (40,000 cars),
- \( \sigma \) is the standard deviation (12,000 cars).

### Step 1: Calculate the Z-score
\[
Z = \frac{46,000 - 40,000}{12,000} = \frac{6,000}{12,000} = 0.5
\]

### Step 2: Find the probability corresponding to the Z-score
We look up the Z-score of 0.5 in the standard normal distribution table, or use a calculator to find the cumulative probability.

The cumulative probability for a Z-score of 0.5 is approximately 0.6915.

### Conclusion
The probability that a randomly selected weekly demand for GM cars is less than 46,000 is approximately **0.6915** or **69.15%**.

Would you like a detailed explanation or any clarifications?

Here are 5 related questions to expand on this topic:

1. What is the probability that the weekly demand is greater than 46,000 cars?
2. What is the probability that the weekly demand is between 34,000 and 46,000 cars?
3. How would the probability change if the standard deviation increased to 15,000 cars?
4. What weekly demand corresponds to the 90th percentile?
5. What is the probability that the weekly demand is less than 30,000 cars?

**Tip:** When dealing with normal distributions, always visualize the problem using the bell curve to better understand probabilities and areas under the curve.

What is the probability that a randomly selected weekly demand for GM is less than 46,000 cars if the demand follows a normal distribution with a mean of 40,000 cars and a standard deviation of 12,000 cars?

Learn how to calculate the probability that the weekly demand for GM cars is less than 46,000 using the normal distribution. This example uses a mean of 40,000 cars and a standard deviation of 12,000 cars, demonstrating how to apply the Z-score formula and interpret standard normal distribution tables. Suitable for high school students in Grades 10-12.

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