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Chapter 4: Problem 14

Spray paint, Part 1. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. Suppose also that you buy three cans of spray paint. (a) How much area would you expect to cover with these three cans of spray paint? (b) What is the standard deviation of the area you expect to cover with these three eans of spray paint? (c) The area you wanted to cover is 80 square feet. What is the probability that you will be able to cover this entire area with these three cans of spray paint?

Short Answer

Expert verified
(a) 75 sq ft, (b) 5.196 sq ft, (c) Probability is 83.15%.

Step by step solution

01

Calculate the Expected Area for One Can

Given that the average area covered by one can is 25 square feet, to find the total expected area covered by three cans, multiply the expected area of one can by the number of cans: \( 25 \times 3 = 75 \) square feet.
02

Calculate the Standard Deviation for Three Cans

The standard deviation for the total area covered by three cans is calculated by multiplying the standard deviation of one can by the square root of the number of cans. Thus, it is \( 3 \times \sqrt{3} = 3 \times 1.732 = 5.196 \) square feet.
03

Find the Z-score for 80 Square Feet

To find the probability of covering 80 square feet, first calculate the z-score using the formula: \( z = \frac{{X - \mu}}{{\sigma}} \), where \( X = 80 \), \( \mu = 75 \), and \( \sigma = 5.196 \). Thus, \( z = \frac{{80 - 75}}{{5.196}} = 0.963 \).
04

Look Up the Z-score in the Normal Distribution Table

Find the probability that corresponds to the z-score of 0.963 in the standard normal distribution table. This probability is approximately 0.8315, which means there is an 83.15% chance that you can cover 80 square feet with three cans.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value of a random variable is like a long-term average. Think of it as what you would anticipate as an outcome if you could repeat an experiment countless times. Here, the expected value concept helps us determine how much area three cans of spray paint can cover on average.
To calculate the expected value, we multiply the expected area for one can by the total number of cans. If one can covers an average of 25 square feet, three cans would cover:
  • Expected area = 25 square feet/can 脳 3 cans = 75 square feet.
This sum tells us that, on average, you would cover 75 square feet with three cans. Each can contributes equally to this expected total. It's a prediction derived from the mean of our distribution.
Standard Deviation
Standard deviation is a measure that tells us how much variability exists in a set of data. It shows how spread out the values are from the mean. A smaller standard deviation indicates data points are close to the mean, while a larger one shows more spread.
For three cans of spray paint, the standard deviation helps us understand how the total area covered can differ from the expected average. We calculate it by multiplying the standard deviation of one can by the square root of the number of cans:
  • Standard deviation for one can = 3 square feet.
  • Total standard deviation = 3 脳 \( \sqrt{3} \) 鈮� 5.196 square feet.
This calculation tells us that while the average coverage is 75 square feet, it can typically vary by about 5.196 square feet from this average.
Z-score
Z-scores allow us to understand a single score's position relative to a group of scores, specifically in a standard normal distribution. It tells us how many standard deviations an element is from the mean.
In the context of spray paint coverage, if we want to find out the likelihood of covering 80 square feet with three cans, we first find the z-score:
  • Formula: \( z = \frac{{X - \mu}}{{\sigma}} \)
  • Where \( X = 80 \) (desired coverage), \( \mu = 75 \) (expected value from our calculation), \( \sigma = 5.196 \) (standard deviation).
  • Thus, \( z = \frac{{80 - 75}}{{5.196}} 鈮� 0.963 \).
This z-score tells us that covering 80 square feet is about 0.963 standard deviations above the expected mean of 75 square feet.
Probability Calculation
Probability calculations allow us to find out the likelihood of an event occurring within a given framework. Using the z-score from the previous section, we determine the probability of covering at least 80 square feet with three cans of spray paint.
Here, for a z-score of 0.963, we look up a standard normal distribution table to find the corresponding probability.
  • A z-score table shows this corresponds to a probability of approximately 0.8315.
  • This means there is an 83.15% chance that three cans of spray paint will cover the desired 80 square feet.
This probability is derived from the standardized normal distribution table and helps us make decisions based on likely outcomes.

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Most popular questions from this chapter

Triathlon times, Part L. In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages \(30-34\) group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 1: 22: 28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is same information on the performance of their groups: \- The finishing times of the Men, Ages \(30-34\) group has a mean of 4313 seconds with a standard deviation of 583 seconds. \- The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. \- The distributions of finishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster finish. (a) What are the Z-ncores for Leo's and Mary's finishing times? What do these Z-scores tell you? (b) Did Leo or Mary rank better in their respective groups? Explain your reasoning. (c) What percent of the triathletes did Leo finish faster than in his group? (d) What percent of the triathletes did Mary finish faster than in her group? (e) If the distributions of finishing times are not nearly normal, would your answers to parts (a)change? Explain your reasoning.

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