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Chapter 6: Problem 46

Cereal A company's single-serving cereal boxes advertise 9.63 ounces of cereal. In fact, the amount of cereal \(X\) in a randomly selected box follows a Normal distribution with a mean of 9.70 ounces and a standard deviation of 0.03 ounces. (a) Let \(Y=\) the excess amount of cereal beyond what's advertised in a randomly selected box, measured in grams ( 1 ounce \(=28.35\) grams). Find the mean and standard deviation of \(Y\). (b) Find the probability of getting at least 3 grams more cereal than advertised. Show your work.

Short Answer

Expert verified
Mean of Y is 1.9845 grams, standard deviation is 0.8505 grams. Probability of 3 grams more is 0.1170.

Step by step solution

01

Define Random Variable Y

Define the random variable \( Y \) as the excess amount of cereal in grams. Since \( X \), the amount of cereal in ounces, is normally distributed, and \( 1 \text{ ounce} = 28.35 \text{ grams} \), we need to convert \( X \) from ounces to grams first: \( Y = (X - 9.63) \times 28.35 \).
02

Calculate Mean of Y

Since \( X \) has a mean of 9.70 ounces, the mean of the excess is \((9.70 - 9.63) = 0.07 \) ounces. Converting this to grams: \( \mu_Y = 0.07 \times 28.35 = 1.9845 \text{ grams} \).
03

Calculate Standard Deviation of Y

The conversion factor from ounces to grams is a constant, so the standard deviation of \( X \) also scales by 28.35. Therefore, \( \sigma_Y = 0.03 \times 28.35 = 0.8505 \text{ grams} \).
04

Find Z-score for 3 Grams Over

We wish to find the probability that \( Y \geq 3 \). First, find the Z-score: \( Z = \frac{3 - 1.9845}{0.8505} \approx 1.193 \).
05

Calculate Probability from Z-score

Using a standard normal distribution table or calculator, find \( P(Z \geq 1.193) \). First, find \( P(Z \leq 1.193) \approx 0.8830 \). Therefore, \( P(Z \geq 1.193) = 1 - 0.8830 = 0.1170 \).
06

Conclusion

The probability of getting at least 3 grams more cereal than advertised is approximately 0.1170.

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Key Concepts

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

Probability Calculation
Probability calculation often involves finding the likelihood that an event will occur given certain conditions. In our example, we're working with a normal distribution, which makes it easier to calculate probabilities. The probability that a random variable takes on a particular value or falls within a certain range can be calculated with the help of the Z-score. Here鈥檚 a simple process:
  • First, understand what you are solving: In the case of cereal boxes, we are interested in finding out the probability that the amount of cereal exceeds the advertised amount by at least 3 grams.
  • Find the Z-score: This step standardizes the problem. The formula for the Z-score is: \( Z = \frac{(X - \mu)}{\sigma} \) where \( X \) is the value you're examining, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • Use the standard normal distribution table: Once you have the Z-score, you can use a table or calculator to find the probability. The table provides the cumulative probability up to the Z-score.
  • Final Probability: In the cereal example, subtract the cumulative probability from 1 to find the probability of Z being greater than the calculated Z-score.
Thus, probability calculations in normal distributions are systematic and rely on understanding the normal curve, Z-scores, and using statistical tables or software accordingly.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It gives us an idea of how much individual values differ from the mean. In the context of a normal distribution, it defines the spread of the distribution curve.
  • Every distribution has its standard deviation, often denoted as \( \sigma \). In our exercise, the standard deviation is given as 0.03 ounces for the amount of cereal in ounces.
  • When converting units, it is essential to adjust the standard deviation by the same conversion factor. For example, ounces to grams is multiplied by 28.35.
  • Standard deviation tells us how much variation we can expect from the average value. A small standard deviation indicates that the values tend to be close to the mean, while a larger standard deviation indicates a wider range of values.
Understanding standard deviation helps in assessing the reliability of average results and understanding what constitutes an unusual value.
Mean of a Random Variable
The mean of a random variable, often symbolized as \( \mu \), represents the average outcome if you were to perform an experiment or gather data repeatedly an infinite number of times. It's the central point in the normal distribution.
  • The mean is crucial because it serves as the balance point of the distribution curve. In the cereal box example, the mean amount of cereal \( X \) is 9.70 ounces.
  • To find the mean in different units or conditions, it involves simple arithmetic. For the excess amount \( Y \), you subtract the advertised mean from the actual mean, then convert to the preferred unit. This gives \( \mu_Y = 0.07 \times 28.35 = 1.9845 \) grams.
  • The mean provides the expected value, showing what happens on average; however, due to natural variations, individual outcomes may differ.
Grasping the concept of mean helps in predicting outcomes and understanding the average trend in data, a foundation for probability calculations and statistical analysis.

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Most popular questions from this chapter

Binomial setting? A binomial distribution will be approximately correct as a model for one of these two settings and not for the other. Explain why by briefly discussing both settings. (a) When an opinion poll calls residential telephone numbers at random, only \(20 \%\) of the calls reach a person. You watch the random digit dialing machine make 15 calls. \(X\) is the number that reach a person. (b) When an opinion poll calls residential telephone numbers at random, only \(20 \%\) of the calls reach a live person. You watch the random digit dialing machine make calls. \(Y\) is the number of calls needed to reach a live person.

Statistics for investing (3.1) Joe's retirement plan invests in stocks through an "index fund" that follows the behavior of the stock market as a whole, as measured by the Standard \(\&\) Poor's \((\mathrm{S} \& \mathrm{P}) 500\) stock index. Joe wants to buy a mutual fund that does not track the index closely. He reads that monthly returns from Fidelity Technology Fund have correlation \(r=0.77\) with the S\&P 500 index and that Fidelity Real Estate Fund has correlation \(r=0.37\) with the index. (a) Which of these funds has the closer relationship to returns from the stock market as a whole? How do you know? (b) Does the information given tell Joe anything about which fund has had higher returns?

A fastfood restaurant runs a promotion in which certain food items come with game pieces. According to the restaurant, 1 in 4 game pieces is a winner. 103\. If Jeff keeps playing until he wins a prize, what is the probability that he has to play the game exactly 5 times? (a) \((0.25)^{5}\) (b) \((0.75)^{4}\) (c) \((0.75)^{5}\) (d) \((0.75)^{4}(0.25)\) (e) \(\left(\begin{array}{l}5 \\ 1\end{array}\right)(0.75)^{4}(0.25)\)

Running a mile A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean 7.11 minutes and standard deviation 0.74 minute. \({ }^{10}\) Choose a student at random from this group and call his time for the mile \(Y\). Find \(P(Y<6)\) and interpret the result.

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