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Using \(N(1152,84)\), the Normal model for weights of Angus steers in Exercise 17, what percent of steers weigh a) over 1250 pounds? b) under 1200 pounds? c) between 1000 and 1100 pounds?

Short Answer

Expert verified
a) 12.17%; b) 71.64%; c) 23.30%

Step by step solution

01

Identify Parameters

First, note that the normal distribution given is \(N(1152, 84)\), where 1152 is the mean (\(\mu\)) and 84 is the standard deviation (\(\sigma\)).For these questions, we need to find the probability of the weights beyond certain limits, which will be translated into \(z\)-scores.
02

Calculate Z-score for 1250 (Part a)

Use the formula for the \(z\)-score: \[ z = \frac{x - \mu}{\sigma} \]For a weight of 1250 pounds, the \(z\)-score is:\[ z = \frac{1250 - 1152}{84} \approx 1.1679 \]
03

Find Probability for Over 1250 Pounds (Part a)

Using a standard normal distribution table or calculator, find the probability for \(z = 1.1679\). This gives the probability of a steer weighing less than 1250 pounds.The table provides \(P(Z < 1.1679) \approx 0.8783\), thus the probability of weighing over 1250 is:\[ P(X > 1250) = 1 - 0.8783 = 0.1217 \] So, 12.17% of steers weigh over 1250 pounds.
04

Calculate Z-score for 1200 (Part b)

Use the \(z\)-score formula:\[ z = \frac{1200 - 1152}{84} \approx 0.5714 \]
05

Find Probability for Under 1200 Pounds (Part b)

Using the standard normal distribution table or calculator, we find \(P(Z < 0.5714)\).This probability is approximately \(0.7164\), so:\[ P(X < 1200) = 0.7164 \] Therefore, 71.64% of steers weigh under 1200 pounds.
06

Calculate Z-scores for 1000 and 1100 (Part c)

Find the \(z\)-scores for 1000 and 1100 pounds:For 1000:\[ z = \frac{1000 - 1152}{84} \approx -1.8095 \]For 1100:\[ z = \frac{1100 - 1152}{84} \approx -0.6190 \]
07

Find Probability Between 1000 and 1100 Pounds (Part c)

Using the standard normal distribution table or calculator:- \(P(Z < -1.8095) \approx 0.0351 \) - \(P(Z < -0.6190) \approx 0.2681 \)To find the probability between 1000 and 1100 pounds:\[ P(1000 < X < 1100) = P(Z < -0.6190) - P(Z < -1.8095) = 0.2681 - 0.0351 = 0.2330 \] Thus, 23.30% of steers weigh between 1000 and 1100 pounds.

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Key Concepts

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

Z-score calculation
Z-score calculation is a crucial part of working with normal distributions in statistics. It helps us understand how far away a particular value is from the mean, in terms of standard deviations. The formula for calculating a z-score is: \[ z = \frac{x - \mu}{\sigma} \] where:
  • \(x\) is the value we are analyzing (e.g., a steer weighing 1250 pounds).
  • \(\mu\) is the mean (average) of the distribution.
  • \(\sigma\) is the standard deviation, which measures the spread of the distribution.
By plugging the numbers into the formula, we can determine how typically a certain observation fits within a distribution. For instance, in our example, a steer weighing 1250 pounds translates to a z-score of approximately 1.1679. This indicates that 1250 pounds is 1.1679 standard deviations above the mean weight of 1152 pounds.
Probability
Probability in the context of normal distribution is about finding the likelihood of a particular value or range of values occurring. Once we have a z-score, we can use standard normal distribution tables or calculators to find probabilities associated with those z-scores. For example:
  • A z-score of 1.1679 has a cumulative probability \(P(Z < 1.1679)\) of approximately 0.8783. This means there's an 87.83% chance that the weight of a steer is less than 1250 pounds.
  • To find the probability of a steer weighing more than 1250 pounds, we calculate \(1 - 0.8783 = 0.1217\) or 12.17%.
  • Similarly, we can determine the probability of a steer weighing between 1000 and 1100 pounds by calculating the cumulative probabilities for the lower and upper z-scores and then finding their difference.
Using these methods, probabilities help us understand real-world scenarios, such as the likelihood of certain conditions in specified ranges.
Statistics
Statistics involves collecting, analyzing, interpreting, and presenting data. Understanding statistics, especially concepts like normal distribution and z-scores, allows us to make informed decisions. Through statistics, we can predict outcomes, compare different datasets, and understand trends or patterns over time. The normal distribution, often called the "bell curve," is particularly useful in statistics because:
  • It describes how data points are spread around a mean in many real-world datasets.
  • The properties of the normal distribution make it easy to determine probabilities and outcomes using z-scores.
  • It applies to various fields such as economics, biology, and social sciences, illustrating its versatility.
By understanding statistics and these core principles, students can critically evaluate data and apply statistical methods to explore various scientific questions and everyday problems.

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Most popular questions from this chapter

Each year thousands of high school students take either the SAT or the ACT, standardized tests used in the college admissions process. Combined SAT Math and Verbal scores go as high as 1600, while the maximum ACT composite score is 36 . Since the two exams use very different scales, comparisons of performance are difficult. A convenient rule of thumb is \(S A T=40 \times A C T+150\); that is, multiply an ACT score by 40 and add 150 points to estimate the equivalent SAT score. An admissions officer reported the following statistics about the ACT scores of 2355 students who applied to her college one year. Find the summaries of equivalent SAT scores. Lowest score \(=19 \quad\) Mean \(=27\) Standard deviation \(=3\) \(\mathrm{Q} 3=30\) Median \(=28 \quad \mathrm{IQR}=6\)

. 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.

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

A company that manufactures rivets believes the shear strength (in pounds) is modeled by \(N(800,50)\). a) Draw and label the Normal model. b) Would it be safe to use these rivets in a situation requiring a shear strength of 750 pounds? Explain. c) About what percent of these rivets would you expect to fall below 900 pounds? d) Rivets are used in a variety of applications with varying shear strength requirements. What is the maximum shear strength for which you would feel comfortable approving this company's rivets? Explain your reasoning.

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of \(188 \mathrm{mg} / \mathrm{dL}\) and a standard deviation of \(24 .\) a) Draw and label the Normal model. b) What percent of adult women do you expect to have cholesterol levels over \(200 \mathrm{mg} / \mathrm{dL}\) ? c) What percent of adult women do you expect to have cholesterol levels between 150 and \(170 \mathrm{mg} / \mathrm{dL}\) ? d) Estimate the IQR of the cholesterol levels. e) Above what value are the highest \(15 \%\) of women's cholesterol levels?

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