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Fruit flies. The common fruit fly Drosophila melanogaster is the most studied organism in genetic research because it is small, is easy to grow, and reproduces rapidly. The length of the thorax (where the wings and legs attach) in a population of male fruit flies is approximately Normal with mean \(0.800\) millimeter (mm) and standard deviation \(0.078 \mathrm{~mm}\). (a) What proportion of flies have thorax length less than \(0.7 \mathrm{~mm}\) ? (b) What proportion have thorax length greater than \(1.0 \mathrm{~mm}\) ? (c) What proportion have thorax length between \(0.7 \mathrm{~mm}\) and \(1.0 \mathrm{~mm}\) ?

Short Answer

Expert verified
(a) 0.1003, (b) 0.0052, (c) 0.8945

Step by step solution

01

Define the normal distribution parameters

The thorax lengths of the fruit flies are normally distributed with a mean \( \mu = 0.800 \) mm and a standard deviation \( \sigma = 0.078 \) mm.
02

Calculate Z-Score for 0.7 mm

To find the proportion of flies with thorax length less than 0.7 mm, calculate the Z-score using the formula:\[ Z = \frac{X - \mu}{\sigma} \] where \(X\) is 0.7 mm, \(\mu\) is 0.800 mm, and \(\sigma\) is 0.078 mm.\[ Z = \frac{0.7 - 0.800}{0.078} \approx -1.282 \]
03

Find Proportion Below 0.7 mm

The Z-score of approximately -1.282 corresponds to a proportion of around 0.1003 from a standard normal distribution table or calculator. Therefore, the proportion of flies with thorax less than 0.7 mm is approximately 0.1003.
04

Calculate Z-Score for 1.0 mm

To find the proportion of flies with thorax length greater than 1.0 mm, calculate the Z-score as follows:\[ Z = \frac{1.0 - 0.800}{0.078} \approx 2.564 \]
05

Find Proportion Above 1.0 mm

The Z-score of approximately 2.564 corresponds to a proportion of around 0.9948.Therefore, the proportion greater than 1.0 mm is:\[ 1 - 0.9948 = 0.0052 \]
06

Calculate Proportion Between 0.7 mm and 1.0 mm

The proportion between 0.7 mm and 1.0 mm is the difference between the proportions found in previous steps:\[ 0.9948 - 0.1003 = 0.8945 \]

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Key Concepts

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

Understanding the Z-Score
The Z-score is a statistical measure that tells us how far a given data point is from the mean of a distribution, expressed in terms of standard deviations. It allows us to understand and compare different data points within a normal distribution.
To calculate the Z-score, we use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
This formula essentially transforms the scale of the data to how many "standard deviations away" it is from the mean. A Z-score of 0 indicates that the data point is exactly at the mean, while a positive or negative score shows how far and in what direction it deviates.
This is crucial in determining probabilities in a normal distribution and is a foundational step in problems involving proportions.
Grasping Standard Deviation
Standard deviation is a key concept in statistics as it measures the amount of variability or dispersion in a set of data. It tells us how much the individual data points differ from the mean of the data set.
In simpler terms, a smaller standard deviation indicates that the data points tend to be close to the mean, while a larger standard deviation indicates that they are spread out over a broader range of values.
To calculate standard deviation, follow these steps:
  • Find the mean of the dataset.
  • Subtract the mean from each data point to find the deviation of each point.
  • Square each deviation.
  • Average these squared deviations.
  • Take the square root of this average.
In the context of the fruit fly problem, the standard deviation gives us insight into how varied the thorax lengths are in the population. Understanding this allows us to use the Z-score effectively to calculate other proportions.
Calculating Proportions in a Normal Distribution
Proportion calculations in a normal distribution tell us what percentage of the data falls above, below, or between certain values. It's a powerful way to make sense of where a particular value stands in a statistical context.
To determine these proportions, we often use Z-scores and refer to the standard normal distribution table, which provides probabilities associated with specific Z-scores.
Here's how it works:
  • Calculate the Z-score for the values of interest using the Z-score formula.
  • Look up the Z-score on the standard normal distribution table to find the probability for that score.
  • For probabilities above a value, subtract from 1, as Z-scores provide the area to the left of a value.
  • For a range, find the probabilities for both Z-scores and subtract them.
For example, in the fruit fly exercise, we calculate proportions of thorax lengths using Z-scores to understand what percentage of flies have lengths below, above, or between certain thresholds. This gives us detailed insights into the distribution of data within the population.

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Most popular questions from this chapter

The Medical College Admissions Test. Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). \({ }^{8}\) A new version of the exam was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 , the mean score was \(500.0\) with a standard deviation of \(10.6\). (a) What proportion of students taking the MCAT had a score over 510 ? (b) What proportion had scores between 505 and 515 ?

Acid rain? Emissions of sulfur dioxide by industry set off chemical changes in the atmosphere that result in "acid rain." The acidity of liquids is measured by \(\mathrm{pH}\) on a scale of 0 to 14 . Distilled water has \(\mathrm{pH} 7.0\), and lower \(\mathrm{pH}\) values indicate acidity. Normal rain is somewhat acidic, so acid rain is sometimes defined as rainfall with a pH below \(5.0\). The \(\mathrm{pH}\) of rain at one location varies among rainy days according to a Normal distribution with mean \(5.43\) and standard deviation \(0.54\). What proportion of rainy days have rainfall with \(\mathrm{pH}\) below \(5.0\) ?

Are we getting smarter? When the Stanford-Binet IQ test came into use in 1932 , it was adjusted so that scores for each age group of children followed roughly the Normal distribution with mean 100 and standard deviation 15 . The test is readjusted from time to time to keep the mean at 100 . If present-day American children took the 1932 Stanford-Binet test, their mean score would be about 120 . The reasons for the increase in IQ over time are not known but probably include better childhood nutrition and more experience in taking tests. 11 (a) IQ scores above 130 are often called "very superior." What percentage of children had very superior scores in 1932 ? (b) If present-day children took the 1932 test, what percentage would have very superior scores? (Assume that the standard deviation 15 does not change.)

Use the Normal Table. Use Table A to find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. (a) \(z<-0.42\) (b) \(z>-1.58\) (c) \(z<2.12\) (d) \(-0.42

The Medical College Admissions Test. A new version of the Medical College Admissions Test (MCAT) was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 , the mean score was \(500.0\) with a standard deviation of \(10.6\) (a) What are the median and the first and third quartiles of the MCAT scores? What is the interquartile range? (b) Give the interval that contains the central \(80 \%\) of the MCAT scores.

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