/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 374 A sample of 25 female Drosophila... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of 25 female Drosophila were measured for number of eggs laid in a 24-hour period, and the following counts were recorded: \begin{tabular}{|c|c|c|c|c|} \hline 97 & 79 & 88 & 91 & 88 \\ \hline 86 & 104 & 89 & 91 & 92 \\ \hline 103 & 96 & 101 & 94 & 91 \\ \hline 92 & 94 & 95 & 88 & 89 \\ \hline 90 & 93 & 84 & 87 & 93 \\ \hline \end{tabular} Calculate for this sample: (a) mean number of eggs laid; (b) variance and standard deviation; (c) standard error of the mean.

Short Answer

Expert verified
(a) The mean number of eggs laid is \(92.64\). (b) The variance and standard deviation can be calculated using the formulas provided in Step 2. (c) The standard error of the mean can be calculated using the formula provided in Step 3.

Step by step solution

01

Compute the mean of the data

To compute the mean (average) number of eggs laid, we first need to find the sum total of the eggs laid and then divide it by the number of samples (25). The total number of eggs laid by the 25 female Drosophila is: \( total = 97+79+88+91+88+86+104+89+91+92+103+96+101+94+91+92+94+95+88+89+90+93+84+87+93 \) Calculating the total, we have: \( total = 2316 \) Divide the total sum by the number of samples (25) to find the mean: \( mean = \frac{total}{Sample \, Size} = \frac{2316}{25} \) Calculating the mean: \( mean = 92.64 \)
02

Calculate variance and standard deviation

First, we need to find the deviation of each data point from the mean and square the deviations. \( (x_i - mean)^2 = (97-92.64)^2, (79-92.64)^2, \dots , (93-92.64)^2 \) Calculate the sum of all squared deviations: \(sum \, of \, squared \, deviations = \sum_{i=1}^{25}(x_i - mean)^2 \) Now, we can calculate the variance: \(variance = \frac{sum \, of \, squared \, deviations}{Sample \, Size-1} \) Calculating the variance: \( variance = \frac{\sum_{i=1}^{25}(x_i - mean)^2}{24} \) Now, we can calculate the standard deviation as the square root of the variance: \(standard \, deviation = \sqrt{variance} \)
03

Calculate the standard error of the mean

To calculate the standard error of the mean (SEM), we use the formula: \( SEM = \frac{standard \, deviation}{\sqrt{Sample \, Size}} \) Calculating the standard error of the mean: \( SEM = \frac{standard \, deviation}{\sqrt{25}} \) Let's summarize our results: (a) The mean number of eggs laid is 92.64. (b) The variance and standard deviation of the dataset can be found using the formulas provided in Step 2. (c) The standard error of the mean can be found using the formula provided in Step 3.

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Key Concepts

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

Data Analysis in Genetics
When studying genetics, data analysis becomes a powerful tool for understanding the inherited characteristics of organisms. By examining traits, such as the egg-laying capability of Drosophila, researchers collect quantitative data which needs to be precisely analyzed to draw meaningful conclusions.

Data analysis in genetics often involves computing descriptive statistics like the mean, which represents the average trait value for a sample population. In the exercise above, the mean number of eggs laid by sample Drosophila females provides insight into the reproductive potential of the population.

Furthermore, variability in genetic data is just as important. Variance and standard deviation are calculated to understand the spread or dispersion around the mean. High variance indicates a wide spread of individual's trait values, pointing to heterogeneity, possibly caused by genetic differences, environmental factors, or both. In the case of our Drosophila population, understanding the variability of egg-laying rates can suggest how uniform the population's genetic factors influencing this trait are.
Statistical Concepts in Biology
Statistical methods are integral to biological research, as biological data tend to be variable and often complex. Through statistical tests and measures such as mean, variance, standard deviation, and standard error, biologists can make sense of data samples and infer characteristics about the entire population.

In the problem related to Drosophila egg-laying, the standard error of the mean specifically is a crucial statistic. It estimates how much the sample mean is likely to vary from the true population mean and provides a measure of the precision of the sample mean. A smaller standard error indicates that the sample mean is a more accurate reflection of the actual population mean, which is extremely valuable when drawing conclusions from a sample to apply to a broader context.

Basic statistical literacy is therefore essential for students of biology, as it underpins the reliable interpretation of experimental data and ensures that biological research conclusions are drawn from a solid foundation of quantifiable evidence.
Drosophila as a Model Organism
Drosophila melanogaster, commonly known as the fruit fly, is a staple model organism in biological research, particularly in genetics. One of the reasons researchers use Drosophila is their rapid life cycle and high rate of reproduction, which allows for the observation of several generations in a relatively short period. This makes them an excellent subject for statistical analysis in experiments like the one featured here.

Moreover, many genetic mechanisms are conserved between Drosophila and other species, including humans. This means that understanding genetic functions in Drosophila can often provide insights into those same functions in more complex organisms. By collecting statistical data on traits such as egg-laying rates, scientists can make inferences about the genetic architecture and evolutionary history of these traits.

The textbook exercise with the egg-laying data is an excellent example of practical applications for biology students learning about inheritance, variability, and statistical analysis in genetics. By interpreting real data from a model organism, students can enhance their understanding of both genetics and statistical analysis in live biological systems, thus gaining a more comprehensive grasp of the core principles of biology.

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Most popular questions from this chapter

The mean spread of a large group of guayule plants is 12 inches. The standard deviation is 2 inches. If the heights are normally distributed, find the probability that a plant picked at random from the group will have a spread: (a) between 10 and 14 inches; (b) greater than 16 inches; (c) of 12 inches. (Assume that heights are recorded to the nearest inch.)

If a random variable \(\mathrm{X}\) is normally distributed with a mean of 118 and a standard deviation of 11 , what \(Z\) -scores correspond to raw scores of 115,134, and \(99 ?\)

In one month at a public health clinic, 70 patients were found to have the same inherited disorder; 55 were males and 15 were females. Assuming a \(1: 1\) ratio for males versus females, use the chi-square \(\left(\mathrm{X}^{2}\right)\) test to determine if any association of the disease with the male sex is statistically significant.

According to early studies (performed by Wright) on guinea pig coat color, the total variance in a randomly mating strain was \(0.573\). The variance in an inbred strain was \(0.340\). From correlations between relatives in the randomly mating population the heritability (gene effect) was determined to be 38 percent of the total variance. Find the following: (a) variance due to gene effect; (b) variance due to epistasis and dominance.

Miniature poodles are thought to have a mean height of 12 inches and a standard deviation of \(1.8\) inches. If height is measured to the nearest inch, find the percentage of poodles having a height exceeding 14 inches.

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