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When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 929 peas, with 705 of them having red flowers. If we assume, as Mendel did, that under these circumstances, there is a \(3 / 4\) probability that a pea will have a red flower, we would expect that \(696.75\) (or about 697 ) of the peas would have red flowers, so the result of 705 peas with red flowers is more than expected. a. If Mendel's assumed probability is correct, find the probability of getting 705 or more peas with red flowers. b. Is 705 peas with red flowers significantly high? c. What do these results suggest about Mendel's assumption that \(3 / 4\) of peas will have red flowers?

Short Answer

Expert verified
The probability of getting 705 or more peas with red flowers is 0.266. This result is not significantly high, which supports Mendel's assumption.

Step by step solution

01

State the given information

There are 929 peas in total, and 705 of them have red flowers. The assumed probability that a pea has a red flower is \( \frac{3}{4} \). We are to find the probability of getting 705 or more peas with red flowers.
02

Define the random variable and its distribution

Let X be the number of peas with red flowers. X follows a binomial distribution with parameters n = 929 (number of trials) and p = \( \frac{3}{4} \) (probability of success, i.e., red flower).
03

Calculate the mean and standard deviation

The mean \( \mu \) and standard deviation \( \sigma \) of a binomial distribution can be calculated as follows:i. Mean: \[ \mu = n \times p \]ii. Standard deviation: \[ \sigma = \sqrt{n \times p \times (1 - p)} \]For this problem:i. \[ \mu = 929 \times \frac{3}{4} = 696.75 \]ii. \[ \sigma = \sqrt{929 \times \frac{3}{4} \times \frac{1}{4}} = \sqrt{929 \times 0.75 \times 0.25} = \sqrt{174.1875} \approx 13.2 \]
04

Use the normal approximation

Since n is large, we can use the normal approximation to the binomial distribution.We convert the binomial random variable to a standard normal random variable using the formula:i. \[ Z = \frac{X - \mu}{\sigma} \]For X = 705:ii. \[ Z = \frac{705 - 696.75}{13.2} \approx \frac{8.25}{13.2} \approx 0.625 \]
05

Find the probability using the standard normal distribution

Using the standard normal distribution table or a calculator, find the probability corresponding to Z = 0.625.i. P(Z < 0.625) \[ \approx 0.734 \]ii. Therefore, the probability of getting Z ≥ 0.625 is:ii. \[ P(Z \geq 0.625) = 1 - P(Z < 0.625) \approx 1 - 0.734 = 0.266 \]
06

Determine if the result is significantly high

To determine if 705 peas with red flowers is significantly high, we typically look for results that are in the top 5% of the distribution (corresponding to a Z-score above 1.645). Since 0.266 (or 26.6%) is not in this range, it is not significantly high.
07

Conclusion on Mendel's assumption

Given that 705 peas with red flowers is not significantly high, the results do not contradict Mendel's assumption that \( \frac{3}{4} \) of peas will have red flowers.

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Key Concepts

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

genetics experiments
In genetics experiments, researchers often study the inheritance of traits through different generations. One of the most notable examples comes from Gregor Mendel, who conducted pioneering work with pea plants. He observed specific ratios of traits, like flower color, across generations and formulated the basic laws of inheritance. For instance, he discovered that when cross-breeding certain pea plants, there was a consistent ratio of plants with red flowers to those with other colors, leading to an expected probability under Mendelian genetics. Understanding these foundational genetics experiments helps us frame problems involving probabilities of inheriting certain traits.
probability calculation
Probability calculation involves determining the likelihood of a specific outcome among all possible outcomes. In Mendel's genetics experiment example, we calculate the probability of a pea having a red flower. If a situation assumes a probability, such as a \( \frac{3}{4} \) chance of red flowers, we can use this probability to predict expected outcomes. The probability of observing a result equal to or greater than a specific number can be obtained by calculating it directly or using statistical tools like the binomial distribution. This step is often the foundation for more complex analyses.
normal approximation
Normal approximation is a technique used to simplify probability calculations for binomial distributions, especially when dealing with a large number of trials. A binomial distribution expresses the number of successes in a fixed number of independent trials, all having the same probability of success. When the number of trials is large, the binomial distribution can be approximated by a normal distribution, which is easier to work with. This involves calculating the mean and standard deviation of the binomial distribution and then using these to form a corresponding normal distribution. This step dramatically simplifies finding cumulative probabilities, which might otherwise require extensive computation or the use of tables.
standard normal distribution
The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It is often denoted as Z. Conversion to the standard normal distribution is done using the formula: \( Z = \frac{X - \mu}{\sigma} \), where X is the value being converted, \mu is the mean, and \sigma is the standard deviation. This conversion allows one to use standard normal distribution tables or a cumulative distribution function calculator to find probabilities. The standard normal distribution is valuable because it standardizes different datasets to a universal form, making it easier to compare and compute probabilities.
statistical significance
Statistical significance is a measure indicating how likely it is that an observed result is due to chance. In the context of Mendel's genetic experiments, determining if having 705 peas with red flowers is significantly high involves comparing the calculated Z-score to a threshold (commonly 1.645 for a 5% significance level). If the Z-score falls within the extreme tail (typically the top or bottom 5%), the result is considered statistically significant, suggesting it is unlikely due to random variation. In Mendel's case, a result not reaching this threshold means 705 peas with red flowers would not be considered significantly high, indicating consistency with the assumed probability model.

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Most popular questions from this chapter

In a study of babies born with very low birth weights, 275 children were given IQ tests at age 8 , and their scores approximated a normal distribution with \(\mu=95.5\) and \(\sigma=16.0\) (based on data from "Neurobehavioral Outcomes of School-age Children Born Extremely Low Birth Weight or Very Preterm," by Anderson et al., Journal of the American Medical Association, Vol. 289, No. 24 ). Fifty of those children are to be randomly selected without replacement for a follow-up study. a. When considering the distribution of the mean IQ scores for samples of 50 children, should \(\sigma_{-}\) be corrected by using the finite population correction factor? Why or why not? What is the value of \(\sigma_{\bar{x}}\) ? b. Find the probability that the mean IQ score of the follow-up sample is between 95 and 105 .

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Use these parameters (based on Data Set 1 "Body Data" in Appendix B): \- Men's heights are normally distributed with mean \(68.6 \mathrm{in.}\) and standard deviation \(2.8 \mathrm{in} .\) \- Women's heights are normally distributed with mean 63.7 in. and standard deviation \(2.9 \mathrm{in}\). The U.S. Navy requires that fighter pilots have heights between 62 in. and 78 in. a. Find the percentage of women meeting the height requirement. Are many women not qualified because they are too short or too tall? b. If the Navy changes the height requirements so that all women are eligible except the shortest \(3 \%\) and the tallest \(3 \%\), what are the new height requirements for women?

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