The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1264 chips and a standard deviation of 116 chips. (a) Determine the 28th percentile for the number of chocolate chips in a bag. (b) Determine the number of chocolate chips in a bag that make up the middle 96% of bag.

How can I solve this with a TI-84?

To solve this problem using a TI-84 calculator, you can utilize the built-in functionality for the normal distribution. Here's how you can find the 28th percentile and the range that makes up the middle 96% of the distribution:

(a) To determine the 28th percentile:
1. Press the "2nd" button, followed by "Vars" button to access the DISTR menu.
2. Scroll down and select "invNorm" (inverse normal), which is option #3.
3. Enter the area to the left of the desired percentile, which in this case is 0.28.
4. Enter the mean (1264) and the standard deviation (116).
5. Press "ENTER" to get the result, which represents the value separating the lower 28% from the upper 72% of the chocolate chips distribution.

(b) To determine the range that makes up the middle 96%:
1. Calculate the Z-scores for the lower and upper percentiles by subtracting and adding half of the desired range to the mean, respectively.
- For the lower percentile: Mean - (Range / 2) = 1264 - (0.96 * Range / 2)
- For the upper percentile: Mean + (Range / 2) = 1264 + (0.96 * Range / 2)
2. Use the Z-scores to find the corresponding values on the number line:
- Use the invNorm function as described in step (a), but now enter 0.5 + (the calculated Z-score) instead of the percentile.
3. The resulting values will be the range within which the middle 96% of the number of chocolate chips in a bag is expected to fall.

By following these steps, you can use your TI-84 calculator to find the desired percentiles and ranges based on the given mean and standard deviation.