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Eggs. A grocery supplier believes that in a dozen eggs, the mean number of broken ones is \(0.6\) with a standard deviation of \(0.5\) eggs. You buy 3 dozen eggs without checking them. a) How many broken eggs do you expect to get? b) What's the standard deviation? c) What assumptions did you have to make about the eggs in order to answer this question?

Short Answer

Expert verified
Expected broken eggs: 1.8; Standard deviation: 0.866; Assumption: Independence.

Step by step solution

01

Understand the Problem

We need to calculate the expected number and standard deviation of broken eggs in 3 dozen, given the mean and standard deviation per dozen.
02

Calculate Expected Number of Broken Eggs

For one dozen eggs, the expected number of broken eggs is given as 0.6. Therefore, for 3 dozen eggs, the total expected number of broken eggs is calculated by multiplying the expected number per dozen by the number of dozens. \[Expected \: Broken \: Eggs = 0.6 \times 3 = 1.8\]
03

Calculate the Standard Deviation for 3 Dozen Eggs

The standard deviation for one dozen eggs is 0.5. Since standard deviations of independent events add up by the square root of the sum of squares, we calculate the combined standard deviation for 3 dozen eggs using the formula:\[Combined \: Standard \: Deviation = \sqrt{0.5^2 \times 3} = \sqrt{0.75} \approx 0.866\]
04

Assumptions

We assume that the number of broken eggs per dozen is independent and follows the same distribution with a mean of 0.6 and a standard deviation of 0.5. This implies each dozen of eggs fails independently with the same probability.

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Key Concepts

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

Expected Value
Expected value is a fundamental concept in probability that represents the average outcome of a random event if it were repeated many times. In the context of our exercise, we're dealing with broken eggs in several dozen. The mean number of broken eggs per dozen is given as 0.6. This is the expected value for a single dozen. To find the expected number for three dozen, we simply multiply the expected value per dozen by the number of dozens:
  • One dozen eggs: 0.6 broken eggs expected
  • Three dozen eggs: 0.6 (per dozen) × 3 (dozens) = 1.8 broken eggs expected
Therefore, when dealing with multiple groups or units, the expected value scales linearly. Always check if your random events, like broken eggs, are independent, as this assumption holds the math together.
Standard Deviation
Standard deviation measures the amount of variation or dispersion in a set of values. It tells you how much the number of broken eggs is likely to differ from the expected value. If the standard deviation is large, the results can vary widely from the expected value.For our problem, the standard deviation for one dozen eggs is 0.5. When considering the standard deviation for three dozen, since the dozen of eggs are independent from each other, we combine them using the square root of the sum of the squares:
  • Standard deviation per dozen: 0.5
  • Three dozen eggs: \[\sqrt{0.5^2 \times 3} = \sqrt{0.75} \approx 0.866\]
This approach ensures we account for all levels of variability, resulting in a better understanding of possible outcomes.
Independent Events
Independence in probability refers to scenarios where the outcome of one event does not affect another. In this exercise, assuming that the number of broken eggs in each dozen is independent is essential. Why? Because if the breaking of eggs was dependent, the calculations for expected value and standard deviation would need a different approach. Understanding independence helps with a few aspects:
  • The calculations remain straightforward, relying on simple multiplication and use of standard formulas.
  • It underpins the correctness of using the square root of the sum of squares for standard deviations.
  • In real-world applications, it allows for more flexible planning and prediction.
When working with any statistical problem, ensure that your events are independent, unless otherwise indicated, to use these simplifying techniques.

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Most popular questions from this chapter

Weightlifting. The Atlas BodyBuilding Company (ABC) sells "starter sets" of barbells that consist of one bar, two 20-pound weights, and four 5 -pound weights. The bars weigh an average of 10 pounds with a standard deviation of \(0.25\) pounds. The weights average the specified amounts, but the standard deviations are \(0.2\) pounds for the 20-pounders and \(0.1\) pounds for the 5 -pounders. We can assume that all the weights are normally distributed. a) \(\mathrm{ABC}\) ships these starter sets to customers in two boxes: The bar goes in one box and the six weights go in another. What's the probability that the total weight in that second box exceeds \(60.5\) pounds? Define your variables clearly and state any assumptions you make. b) It costs \(\mathrm{ABC} \$ 0.40\) per pound to ship the box containing the weights. Because it's an odd-shaped package, though, shipping the bar costs \(\$ 0.50\) a pound plus a \(\$ 6.00\) surcharge. Find the mean and standard deviation of the company's total cost for shipping a starter set. c) Suppose a customer puts a 20-pound weight at one end of the bar and the four 5 -pound weights at the other end. Although he expects the two ends to weigh the same, they might differ slightly. What's the probability the difference is more than a quarter of a pound?

Insurance. An insurance policy costs \(\$ 100\) and will pay policyholders \(\$ 10,000\) if they suffer a major injury (resulting in hospitalization) or \(\$ 3000\) if they suffer a minor injury (resulting in lost time from work). The company estimates that each year 1 in every 2000 policyholders may have a major injury, and 1 in 500 a minor injury only. a) Create a probability model for the profit on a policy. b) What's the company's expected profit on this policy? c) What's the standard deviation?

Pick a card, any card. You draw a card from a deck. If you get a red card, you win nothing. If you get a spade, you win \(\$ 5\). For any club, you win \(\$ 10\) plus an extra \(\$ 20\) for the ace of clubs. a) Create a probability model for the amount you win. b) Find the expected amount you'll win. c) What would you be willing to pay to play this game?

Defects. A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips, etc.). While none had more than three of these defects, \(7 \%\) had three, \(11 \%\) two, and \(21 \%\) one defect. Find the expected number of appearance defects in a new car and the standard deviation.

Garden. A company selling vegetable seeds in packets of 20 estimates that the mean number of seeds that will actually grow is 18 , with a standard deviation of \(1.2\) seeds. You buy 5 different seed packets. a) How many bad seeds do you expect to get? b) What's the standard deviation? c) What assumptions did you make about the seeds? Do you think that assumption is warranted? Explain.

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