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Biological measurements on the same species often follow a Normal distribution quite closely. The weights of seeds of a variety of winged bean are approximately Normal with mean 525 milligrams (mg) and standard deviation 110 mg. (a) What percent of seeds weigh more than 500 mg? Show your method. (b) If we discard the lightest 10% of these seeds, what is the smallest weight among the remaining seeds? Show your method. Exercises 100 and 101 refer to the following setting. A researcher studied a group of identical twins who had been separated and adopted at birth. In each case, one twin (Twin A) was adopted by a low-income family and the other (Twin \(\mathrm{B}\) ) by a high-income family. Both twins were given an \(\mathrm{IQ}\) test as adults. Here are their scores: \(^{47}\) Twin A: 120 99 99 94 111 97 99 94 104 114 113 100 Twin B: 128 104 108 100 116 105 100 100 103 124 114 112

Short Answer

Expert verified
(a) 59.10% of seeds weigh more than 500 mg. (b) Minimum remaining seed weight is 385 mg.

Step by step solution

01

Identify Given Information

For the seed weights problem, we know that the weights follow a normal distribution with a mean (\(\mu\)) of 525 mg and a standard deviation (\(\sigma\)) of 110 mg.
02

Convert Weights to Z-Scores for Part (a)

We need to find the probability that a seed weighs more than 500 mg. First, we calculate the z-score for 500 mg using the formula \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the weight of interest. Thus, \(z = \frac{500 - 525}{110} \approx -0.227\).
03

Find Probability Using Z-Table for Part (a)

Using a z-table or calculator, find the probability that corresponds to \(z = -0.227\). This value is approximately 0.4090, which represents the probability of weighing less than 500 mg. The probability of weighing more is \(1 - 0.4090 = 0.5910\).
04

Interpret Probability for Part (a)

About 59.10% of seeds weigh more than 500 mg.
05

Identify Requirement for Part (b)

We want to find the weight corresponding to the 10th percentile, i.e., the weight below which 10% of the seeds fall. This requires finding the z-score that corresponds to 10% in a standard normal distribution (z-table value of -1.282).
06

Calculate Weight Using Z-Score for Part (b)

Using the z-score, calculate the weight with \(x = \mu + z\sigma\). Substituting in the known values gives \(x = 525 + (-1.282 \times 110) \approx 384.98\) mg.
07

Conclusion for Part (b)

Thus, the smallest weight among the remaining seeds (after discarding the lightest 10%) is approximately 385 mg.

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Key Concepts

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

Understanding Z-Score
The z-score is a simple yet powerful tool used in statistics to determine how many standard deviations an element is from the mean of a distribution. It is particularly useful when dealing with a Normal distribution, which has the classic bell-shaped curve.
To calculate the z-score of a particular data point, we use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
In this formula, \(x\) represents the value we are analyzing, \(\mu\) is the average or mean of the distribution, and \(\sigma\) is the standard deviation. A positive z-score means that the data point is above the average, whereas a negative z-score indicates a value below the average.
This statistic is instrumental because it allows us to compare individual data points from different distributions or even compare different datasets on a standardized scale. Calculating and interpreting the z-score helps you identify how exceptional or typical a specific measurement is within a dataset.
Exploring Percentiles
Percentiles are a key concept in statistics that helps to understand the relative standing of a value within a dataset. In a standard Normal distribution, the percentiles denote the percentage of data points that fall below a particular z-score.
For example, if a seed's weight falls at the 10th percentile, it means that 10% of all seeds are lighter than this seed. To find the weight corresponding to a specific percentile, such as the 10th percentile, one has to first find the z-score that corresponds to that percentile.
Once the z-score is known, you can plug this back into the z-score formula (\( x = \mu + z \sigma \)) to convert the percentile back into an actual weight or value in the original measurement units. Understanding percentiles is essential for interpreting scores and data since they provide insights into how a particular observation compares to the rest of the dataset.
Probability and the Normal Curve
Probability in statistics gives you a measure of how likely an event is to occur. In the context of a Normal distribution, it tells you the likelihood of a particular observation or measurement happening within your data.
The area under the Normal curve represents the total probability, which is always equal to 1. Specific sections of the curve can represent probabilities associated with particular ranges of data.
  • For instance, the probability of a seed weighing more than a certain amount would be the area of the curve to the right of that weight.
  • The probability of a seed weighing less than another value is to the left of that value on the curve.
With the use of a z-table or statistical software, we can easily translate z-scores into probabilities, helping to answer questions about how common or rare a particular data point is within the distribution.
Diving into Standard Deviation
The standard deviation is a crucial statistic that measures the amount of variation or dispersion in a dataset. It tells you how much the individual data points in a dataset differ from the mean, helping to quantify the spread.
A small standard deviation indicates that the observations are close to the mean, whereas a large standard deviation shows that the values are spread out over a wider range.
In the context of a Normal distribution, the standard deviation is pivotal in determining probabilities, z-scores, and percentile positions. Each step away from the mean (measured in standard deviations) provides concrete insight into the distribution's characteristics.
When dealing with Normally distributed data, about 68% of the data falls within one standard deviation of the mean, approximately 95% within two standard deviations, and nearly 99.7% within three standard deviations. Understanding standard deviation equips you to better grasp the variability and predictability inherent in the data.

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