Question

Assuming problems 1 - 4 have data that is normally distributed, answer the following questions:

1. The shelf life of a particular dairy product is normally distributed with a mean of 12 days and a standard deviation of 3 days
a. About what percent of the products last between 10 and 12 days?
b. About what percent of the products last 8 days or less?

2. The mean life of a tire is 30,000 km. The standard deviation is 2000 km.
a. What pereent of the tires will have a life that exceeds 28,000 km ?
b. If a company purchased 2000 tires, how many tires would you expect to last more than 29000 km ?

3. A line up for tickets to a local concert had an average (mean) waiting time of 20 minutes with a standard deviation of 4 minutes.
a. What percentage of the people in line waited for more than 26 minutes?
b. If 2000 ticket buyers were in line, how many of them would expect to wait for less than 14 minutes?

4. On a recent math test, the mean score was 75 and the standard deviation was 5. Mike made 93 . Would his grade be considered unusual if the grades were normally distributed? Explain. What questions do you have for me?

Answer 1

In a series of problems involving normally distributed data, the Z-score is used to calculate the percentage of data within certain intervals. For each scenario, Z-scores provide a standardized way to determine where a particular value lies within the distribution, allowing us to answer questions about shelf life, tire longevity, wait times, and test scores.

Step-by-step explanation:

Normal Distribution Problems

For the dairy product with a shelf life normally distributed with a mean of 12 days and a standard deviation of 3 days, to calculate the percentage of products lasting between 10 and 12 days and those lasting 8 days or less, you would use the Z-score formula Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

For the tire life example, if the mean is 30,000 km and the standard deviation is 2,000 km, you would first find the Z-score for 28,000 km and look up the corresponding percentile in a standard normal distribution table to find the percentage of tires that exceed that lifespan. To determine how many tires out of 2000 would last more than 29,000 km, calculate the Z-score for 29,000 km, determine the corresponding percentile, and then multiply the percentage by the total number of tires.

In the question about the concert line up, to determine the percentage of people waiting more than 26 minutes and how many out of 2000 would wait less than 14 minutes, apply a similar Z-score approach as the previous examples.

For Mike's math test score, first calculate his Z-score with a mean of 75 and a standard deviation of 5. Then compare this Z-score to the typical range to establish if his grade is considered unusual within the normal distribution.