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More cattle Recall that the beef cattle described in Exercise 29 ?had a mean weight of 1152 pounds, with a standard deviation of 84 pounds. a. Cattle buyers hope that yearling Angus steers will weigh at least 1000 pounds. To see how much over (or under) that goal the cattle are, we could subtract 1000 pounds from all the weights. What would the new mean and standard deviation be? b. Suppose such cattle sell at auction for 40 cents a pound. Find the mean and standard deviation of the sale prices (in dollars) for all the steers.

Short Answer

Expert verified
a. The mean weight after subtracting 1000 pounds from all the weights would be 152 pounds and the standard deviation would be 84 pounds. \n b. The mean and standard deviation of the sale prices in dollars for all the steers would be 460.8 dollars and 33.6 dollars respectively.

Step by step solution

01

Calculation for the new mean after data shifting

To find the new mean, we subtract 1000 from the original mean. This is because when we subtract a constant from every value, the mean changes by that same amount.\n\nSo, the new mean = original mean - 1000\n= 1152 - 1000\n= 152 pounds.
02

Calculation for the new standard deviation after data shifting

The standard deviation tells us the average distance between the data values and their mean. Shifting all of the data by the same amount does not change the average distance between the data values and their mean. Therefore, when we shift data, the standard deviation doesn't change.\n\nSo, the new standard deviation = original standard deviation = 84 pounds.
03

Calculation for the new mean after data scaling

To find the new mean after data are scaled (i.e., multiplied by a constant), we multiply the original mean by that constant. This is because when we multiply every value by a constant, the mean changes by the same factor.\n\nSo, new mean = original mean * 0.40\n= 1152 * 0.40\n= 460.8 dollars.
04

Calculation for the new standard deviation after data scaling

The standard deviation gets larger when the values are more spread out. Multiplying all of the data by the same positive number changes the spread of the data by the same factor. Therefore, the new standard deviation is the original standard deviation times the constant.\n\nSo, new standard deviation = original standard deviation * 0.40\n= 84 * 0.40\n= 33.6 dollars.

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Key Concepts

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

Mean Calculation
The mean, or average, of a set of data is one of the most fundamental concepts in statistics. It provides us with a single value that summarizes the entire dataset. A mean is calculated by adding up all the values in a dataset and then dividing by the number of values.In our original exercise, the mean weight of the cattle is 1152 pounds. When data is shifted uniformly, as in subtracting 1000 pounds from each cattle's weight, the mean also shifts by the same amount without affecting the distribution of the data relative to the mean. The mean is sensitive to every value in the dataset, making it a useful tool but also susceptible to outliers. However, in the example given, there are no outliers mentioned, so the mean gives us a good central tendency of the cattle weights.Let's say we have five cattle weighing 1100, 1200, 1150, 1050, and 1200 pounds. To calculate the mean, we sum the weights: 1100 + 1200 + 1150 + 1050 + 1200 = 5700 pounds. Since there are five cattle, we divide by 5, which gives us a mean weight of 1140 pounds. This process illustrates how every value contributes to the overall average.
Standard Deviation
Standard deviation is a measure of dispersion or variability in a dataset. It tells us, on average, how much each data point differs from the mean of the data set. A smaller standard deviation indicates that the data points are closer to the mean, while a larger standard deviation points to data that are more spread out.In this case, the original standard deviation of the cattle weights is 84 pounds. When we shift all weights by subtracting 1000 pounds, the standard deviation remains unchanged at 84 pounds. This is because standard deviation is concerned with the relative distance of each data point from the mean, which does not change when you add or subtract a constant to all values.In practice, calculating standard deviation involves several steps: finding the mean, calculating the difference between each data value and the mean, squaring those differences, averaging those squared differences, and then taking the square root of that average. For instance, if the weights of the cattle were identical to the mean provided, the standard deviation would be zero, indicating no variability among the cattle's weights.
Data Scaling
Data scaling involves adjusting the range of data values using multiplication or division, which alters the scale but not the shape of the data distribution. When you sell cattle by the pound, for example, the price per pound scales the data. If the original mean weight is 1152 pounds and the price is \(0.40 per pound, we multiply the mean weight by the price to find the mean sale price.In data scaling, the new mean is calculated by multiplying the original mean by the scaling factor, resulting in a new mean of \)460.8 in this case. Similarly, the standard deviation is also scaled by the same factor, leading to a new standard deviation of $33.6. It's important to understand that scaling changes the units of the data, but it does not change the overall shape of the distribution; it's simply a change of scale. In the cattle sale example, we're changing the context from weight to sale price, providing useful insights for economic analysis but keeping the statistical characteristics relative to each other.

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Most popular questions from this chapter

Using \(N(1152,84)\), the Normal model for weights of Angus steers in Exercise 13 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?

A popular band on tour played a series of concerts in large venues. They always drew a large crowd, averaging 21,359 fans. While the band did not announce (and probably never calculated) the standard deviation, which of these values do you think is most likely to be correct: \(20,200,2000,\) or 20,000 fans? Explain your choice.

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 \(=19\) Mean \(=27\) Standard deviation \(=3 \mathrm{Q} 3=30\) Median \(=28 \mathrm{IQF}\)

Environmental Protection Agency (EPA) fuel economy estimates for automobile models tested recently predicted a mean of \(24.8 \mathrm{mpg}\) and a standard deviation of \(6.2 \mathrm{mpg}\) for highway driving. Assume that a Normal model can be applied. a. Draw the model for auto fuel economy. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b. In what interval would you expect the central \(68 \%\) of autos to be found? c. About what percent of autos should get more than 31 mpg? d. About what percent of cars should get between 31 and \(37.2 \mathrm{mpg} ?\) e. Describe the gas mileage of the worst \(2.5 \%\) of all cars.

Exercise 10 proposes modeling IQ scores with \(N(100,15)\). What IQ would you consider to be unusually high? Explain.

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