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Chapter 14: Problem 8

An automatic filling machine in a factory fills bottles of ketchup with a mean of 16.1 oz and a standard deviation of 0.05 oz with a distribution that can be well modeled by a Normal model. What is the probability that your bottle of ketchup contains less than 16 oz?

Short Answer

Expert verified
The probability that your bottle of ketchup contains less than 16 oz is 0.0228 or 2.28%.

Step by step solution

01

Identify the parameters of the Normal Distribution

The problem tells that weight of ketchup filled by the machine follows a Normal Distribution. The mean (\( \mu \)) is 16.1 oz and the standard deviation (\( \sigma \)) is 0.05 oz.
02

Calculate the z-score

The Z score is a measure of how many standard deviations an element is from the mean. To find probability that a bottle has less than 16 oz of ketchup, we need to find the z-score of 16. The formula of z-score is \( z = \frac{x - \mu}{\sigma} \), where x is the value that we're finding the probability for. In this case, x = 16, \( \mu \) = 16.1, and \( \sigma \) = 0.05. Plugging in these values, we get \( z = \frac{16 - 16.1}{0.05} = -2 \). This means that 16 oz is 2 standard deviations below the mean.
03

Find the probability corresponding to the z-score

Now, in order to find the probability of having less than 16 oz, look up the probability corresponding to z = -2 on the standard normal distribution table (or use a technology tool). The table value at z = -2 is 0.0228 or 2.28%.

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Key Concepts

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

Understanding the Z-Score
The **z-score** is an important concept in statistics, especially when working with normal distributions. It tells us how many standard deviations a specific value is from the mean. This helps us determine how unusual or typical a particular data point is in a distribution.
To calculate the z-score, you use the formula: \[ z = \frac{x - \mu}{\sigma} \] where:
  • \( x \) is the value in question.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation.

This calculation places the value on the standard normal distribution, which is a special kind of normal distribution with a mean of 0 and a standard deviation of 1. Once you have the z-score, you can find out what percentage of data falls below that value.
The Role of Standard Deviation
**Standard deviation** is a measure of how spread out the values in a data set are. It's a key concept in understanding a normal distribution. When you have a smaller standard deviation, most values are closer to the mean. Conversely, a larger standard deviation indicates more variability in the dataset.
Calculating standard deviation involves several steps:
  • First, find the mean (average).
  • Subtract the mean from each data point and square the result.
  • Find the average of these squared differences.
  • Take the square root of this average, and you have the standard deviation.

In the context of our bottle-filling example, the standard deviation helps us understand how much individual bottle fills deviate from the average amount of 16.1 oz.
Probability and Its Calculation
**Probability** quantifies the likelihood of an event happening and is crucial in statistics. When dealing with a normal distribution, probability helps us find the odds of a particular outcome, like a bottle containing less than 16 oz.
To calculate this probability, we first determine the z-score. Using the z-score, we refer to the standard normal distribution table or use software tools to find the corresponding probability value.
  • The z-score tells us how many standard deviations a value is from the mean.
  • The probability found in the table reflects how much of the distribution falls below this z-score.

In our case, a z-score of -2 corresponds to a probability of 0.0228. This means there's a 2.28% chance a bottle will have less than 16 oz of ketchup.
The Importance of the Mean
The **mean** is one of the most fundamental concepts in statistics. It represents the average value of a dataset. In a normally distributed dataset, the mean is the central point where the highest peak of the bell curve occurs.
To find the mean, you sum up all the data points and divide by the number of points.
  • For our ketchup bottles, a mean of 16.1 oz tells us that on average, each bottle is filled with this amount.
  • A mean is useful for comparing individual data points and understanding their deviation in context.

The mean also plays a critical role in calculating the z-score and analyzing the spread of a distribution, as it's a reference point for measuring variability.

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Most popular questions from this chapter

The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl. a. How much more cereal do you expect to be in the large bowl? b. What's the standard deviation of this difference? c. If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d. What are the mean and standard deviation of the total amount of cereal in the two bowls? e. If the total follows a Normal model, what's the probability you poured out more than 4.5 ounces of cereal in the two bowls together? f. The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the expected amount of cereal left in the box and 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?

Day trading An option to buy a stock is priced at $$\$ 200$$. If the stock closes above 30 on May \(15,\) the option will be worth $$\$ 1000$$. If it closes below \(20,\) the option will be worth nothing, and if it closes between 20 and 30 (inclusively), the option will be worth $$\$ 200$$. A trader thinks there is a \(50 \%\) chance that the stock will close in the \(20-30\) range, a \(20 \%\) chance that it will close above 30 , and a \(30 \%\) chance that it will fall below 20 on May \(15 .\) a. Should she buy the stock option? b. How much does she expect to gain? c. What is the standard deviation of her gain?

A farmer has \(100 \mathrm{lb}\) of apples and \(50 \mathrm{lb}\) of potatoes for sale. The market price for apples (per pound) each day is a random variable with a mean of 0.5 dollars and a standard deviation of 0.2 dollars. Similarly, for a pound of potatoes, the mean price is 0.3 dollars and the standard deviation is 0.1 dollars. It also costs him 2 dollars to bring all the apples and potatoes to the market. The market is busy with eager shoppers, so we can assume that he'll be able to sell all of each type of produce at that day's price. a. Define your random variables, and use them to express the farmer's net income. b. Find the mean. c. Find the standard deviation of the net income. d. Do you need to make any assumptions in calculating the mean? How about the standard deviation?

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. 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?
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