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. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that averages 74 grams, but \(11 \%\) of the tomatoes are too small. It is reasonable to assume that a Normal model applies. a) What is the standard deviation of the weights of Romas now being grown? b) Scientists hope to reduce the frequency of undersized tomatoes to no more than \(4 \%\). One way to accomplish this is to raise the average size of the fruit. If the standard deviation remains the same, what target mean should they have as a goal? c) The researchers produce a new variety with a mean weight of 75 grams, which meets the \(4 \%\) goal. What is the standard deviation of the weights of these new Romas? d) Based on their standard deviations, compare the tomatoes produced by the two varieties.

Short Answer

Expert verified
a) 3.25 grams. b) 75.68 grams. c) 2.86 grams. d) New variety has less variability.

Step by step solution

01

Understanding the Normal Distribution

We know that the weights of Roma tomatoes follow a Normal distribution. This means we can use the properties of the Normal distribution, such as the empirical rule, to find unknowns like mean and standard deviation.
02

Solve for Standard Deviation in Part a

Given that 11% of tomatoes weigh less than 70 grams, we find the z-score corresponding to the cumulative probability of 0.11 using the standard normal table (approximately -1.23). The known mean is 74 grams. Using the z-score formula:\[ z = \frac{70 - 74}{\sigma} = -1.23 \] we solve for \( \sigma \), the standard deviation.\[ \sigma = \frac{4}{1.23} \approx 3.25 \] grams.
03

Solve for Target Mean in Part b

To reduce undersized tomatoes to 4%, find the z-score corresponding to this cumulative probability (approximately -1.75). Set up the z-score formula with the same standard deviation:\[ -1.75 = \frac{70 - \mu}{3.25} \]Solving for \( \mu \), the mean:\[ \mu = 70 + 1.75 \times 3.25 \approx 75.68 \] grams.
04

Solve for Standard Deviation in Part c

Researchers aim for a mean of 75 grams to achieve a 4% undersize rate. Use the z-score corresponding to 4% (-1.75), and solve for the new standard deviation:\[ -1.75 = \frac{70 - 75}{\sigma} \]\[ \sigma = \frac{5}{1.75} \approx 2.86 \] grams.
05

Comparison of Variability in Part d

The original variety has a standard deviation of approximately 3.25 grams, whereas the new variety, which meets the 4% goal, has a standard deviation of approximately 2.86 grams. This indicates that the new variety has less variability, which is desirable as it produces more consistently sized tomatoes.

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Key Concepts

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

Standard Deviation
Standard deviation is an important concept when dealing with the Normal Distribution. Simply put, it measures how much the values in a data set deviate from the mean, or average, value. In the case of Roma tomatoes, it helps us understand the variation in tomato weights.
For example, if the standard deviation is large, the weights of the tomatoes vary widely from the mean. Conversely, a small standard deviation means the weights are more consistently closer to the mean.
This gives us insight into the consistency of the tomato sizes, which is crucial for growers who want to ensure that their produce meets specific market demands, such as minimum weight requirements.
Z-score Calculation
The Z-score is a useful statistic that shows how many standard deviations an element is from the mean. It's calculated using this formula:
  • \[ z = \frac{x - \mu}{\approx} \]
Where:
  • \( z \) is the Z-score
  • \( x \) is the value
  • \( \mu \) is the mean
  • \( \approx \) is the standard deviation
For example, if the Z-score is -1.23, it means the value is 1.23 standard deviations below the mean. In agriculture, Z-scores can help identify how individual measurements compare to the average, playing a significant role in determining the quality of produce like tomatoes based on size and weight.
Agricultural Measurement
Agricultural measurement revolves around quantifying specific attributes, such as plant or produce weight, size, or yield. These specifications are crucial for optimizing growth and ensuring products meet market standards.
In the case of Roma tomatoes, weights are measured to ensure that the tomatoes fall within a preferred range for consumer acceptance. Too small, and they might not sell. Hence, it is crucial to understand the analytics behind these measurements. This can help producers adjust farming practices to achieve more favorable results, like changing the mean or reducing variability in weight distribution.
Mean Calculation
The mean, or average, is a fundamental measure in statistics, representing the typical value in a data set. To find the mean weight of tomatoes, you add up all the tomato weights and divide by the number of tomatoes.
In practice:
  • Gather all individual tomato weights.
  • Add these weights together.
  • Divide by the total number of tomatoes considered.
The mean provides a central value around which the other data points are spread. It's crucial for determining whether the average tomato meets desired weight standards, guiding decisions on whether to alter breeding strategies. For instance, growers may aim to adjust the mean to a more market-friendly weight to meet consumer preferences.

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Most popular questions from this chapter

The first Stat exam had a mean of 80 and a standard deviation of 4 points; the second had a mean of 70 and a standard deviation of 15 points. Reginald scored an 80 on the first test and an 85 on the second. Sara scored an 88 on the first but only a 65 on the second. Although Reginald's total score is higher, Sara feels she should get the higher grade. Explain her point of view.

. Exercise 26 proposes modeling IQ scores with \(N(100,16) .\) What IQ would you consider to be unusually high? Explain.

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The Virginia Cooperative Extension reports that the mean weight of yearling Angus steers is 1152 pounds. Suppose that weights of all such animals can be described by a Normal model with a standard deviation of 84 pounds. a) How many standard deviations from the mean would a steer weighing 1000 pounds be? b) Which would be more unusual, a steer weighing 1000 pounds or one weighing 1250 pounds?

Based on the Normal model \(N(100,16)\) describing IQ scores, what percent of people's IQs would you expect to be a) over \(80 ?\) b) under \(90 ?\) c) between 112 and \(132 ?\)

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