Question

Based on the model \(N(1152,84)\) describing Angus steer weights, what are the cutoff values for a) the highest \(10 \%\) of the weights? b) the lowest \(20 \%\) of the weights? c) the middle \(40 \%\) of the weights?

Short answer

a) 1259.52 lbs b) 1081.44 lbs c) 1108.32 lbs to 1195.68 lbs

Step-by-step solution

Understand the Problem

The problem gives us a normal distribution of steer weights modeled as \(N(1152, 84)\), where 1152 is the mean and 84 is the standard deviation. We need to find cutoff values for specific percentiles of this distribution.

Calculating Cutoff for Highest 10%

To find the weight cutoff for the highest 10% of the distribution, we look for the 90th percentile (since the highest 10% starts above the 90% mark). Use a standard normal distribution table or calculator to find the z-score corresponding to the 90th percentile, which is approximately 1.28. Then, use the formula:\[ x = \mu + z \times \sigma \]Substitute \( \mu = 1152 \), \( \sigma = 84 \), and \( z = 1.28 \):\[ x = 1152 + 1.28 \times 84 = 1259.52 \]Thus, the cutoff weight is approximately 1259.52 lbs.

Calculating Cutoff for Lowest 20%

For the lowest 20%, we need the 20th percentile. The z-score for the 20th percentile is approximately -0.84. Use the formula:\[ x = \mu + z \times \sigma \]With \( \mu = 1152 \), \( \sigma = 84 \), and \( z = -0.84 \):\[ x = 1152 + (-0.84) \times 84 = 1081.44 \]Therefore, the cutoff weight for the lowest 20% is approximately 1081.44 lbs.

Calculating Cutoff for Middle 40%

The middle 40% lies between the 30th percentile and the 70th percentile. First, find the z-scores for the 30th and 70th percentiles, which are approximately -0.52 and 0.52, respectively. Calculate each cutoff:For the 30th percentile:\[ x_{30} = 1152 + (-0.52) \times 84 = 1108.32 \]For the 70th percentile:\[ x_{70} = 1152 + 0.52 \times 84 = 1195.68 \]The cutoff values for the middle 40% are approximately 1108.32 lbs and 1195.68 lbs.

Key concepts

Percentiles

Percentiles are a way to understand how a particular value in a data set compares to the rest of the data. When we talk about a specific percentile, we're referring to the value below which a certain percentage of the data falls. For example, the 90th percentile is the value below which 90% of the observations fall. In our exercise, we calculate percentiles to find out the cutoff weights of Angus steers based on their distribution. To find a particular percentile in a normal distribution, you first determine the corresponding z-score using a z-table or calculator:

• For the highest 10%: We look for the 90th percentile, which corresponds to a z-score of about 1.28.
• For the lowest 20%: We identify the 20th percentile, linked to a z-score of approximately -0.84.
• The middle 40% involves percentiles from 30th to 70th, and involves z-scores of around -0.52 for the 30th and 0.52 for the 70th.

Percentiles help us understand the position of a data point relative to the entire distribution. Each calculation varies based on the specific subset of data you're analyzing.

Z-score

The z-score is an essential concept when working with normal distributions. It represents the number of standard deviations a data point is from the mean. The formula for a z-score is:\[ z = \frac{x - \mu}{\sigma} \]Here, \(x\) is the individual data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.In many problems, and particularly in this exercise, the z-score allows us to find how far a particular value is from the mean of a normal distribution. Once we know the z-score, we can easily determine the corresponding percentile and vice versa:

• A positive z-score (e.g., 1.28 for the 90th percentile) indicates a data point is above the mean.
• A negative z-score (e.g., -0.84 for the 20th percentile) signifies a point below the mean.

Calculating z-scores is a fundamental step when identifying specific data points within a normal distribution. This calculation allows for standardizing data to easily compare different datasets.

Standard Deviation

Standard deviation is a measure of how spread out numbers are in a dataset. It tells us how much individual data points deviate from the mean on average. In the context of a normal distribution like in our Angus steer model, standard deviation helps in determining how much the weights vary around the *average* weight of 1152 lbs.In our exercise, the standard deviation is 84 lbs. This value helps us compute the z-score and eventually find the percentiles. Here's why standard deviation is essential:

• Larger standard deviations indicate data points are spread out over a broader range. In contrast, smaller deviations suggest data points are closer to the mean.
• When we use the formula \( x = \mu + z \times \sigma \), the standard deviation (\(\sigma = 84\)) acts as a scaling factor, shifting the z-score away from or towards the mean.

A firm grasp of standard deviation allows students to understand how consistent or varied their data points within any given distribution.