The weight of bags of onion are normally distributed with a mean of 32

oz, and standard d of 0.36 oz.bgs in the upper 4.5% are too heavy and must be repackaged.what is the most a bag of onions can weight and not be repackaged.i don't know how to solve. am not asking the tutors to do my work
am just asking how should i deal with
this particular question.

Consult your Z table to see how many std away leaves 4.5%

Multiply that by 0.36 and add it to the mean.

Play around with Z table stuff at

http://davidmlane.com/hyperstat/z_table.html

To solve this question, you need to find the weight at which 4.5% of the bags are heavier. Here's how you can approach it:

1. Begin by understanding the properties of a normal distribution. Since the weights of bags of onions are normally distributed, you can use the properties of the standard normal distribution to solve this problem.

2. Identify the z-score associated with the upper 4.5% of the distribution. The z-score gives you a standardized value for any observation in a normal distribution. To find the z-score, you can use a standard normal distribution table or use statistical software.

3. In this case, 4.5% of the distribution is in the upper tail, so you need to find the z-score that corresponds to an area of 0.045 (1 - 0.045 = 0.955 for the lower 95.5%). The z-score can be found using a standard normal distribution table or a calculator with a built-in function for finding the z-score corresponding to a given area.

4. Once you have the z-score, you can convert it back to the original measurement scale of ounces by using the formula:
x = z * σ + μ
where x is the weight in ounces, z is the z-score, σ is the standard deviation (0.36 oz in this case), and μ is the mean (32 oz).

5. Plug in the values to find the weight at which 4.5% of the bags are heavier:
x = z * σ + μ
x = z * 0.36 + 32

By following these steps, you can determine the maximum weight a bag of onions can have without needing to be repackaged.