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Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 6, Problem 6.3.18a

Hybridization A hybridization experiment begins with four peas having yellow pods and one pea having a green pod. Two of the peas are randomly selected with replacement from this population.


a. After identifying the 25 different possible samples, find the proportion of peas with yellow pods in each of them, then construct a table to des

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Define the population: The population consists of 4 peas with yellow pods and 1 pea with a green pod. Since the selection is with replacement, the probability of selecting a yellow pod pea is \( P(Y) = \frac{4}{5} \) and the probability of selecting a green pod pea is \( P(G) = \frac{1}{5} \).
List all possible samples: Since two peas are selected with replacement, there are \( 5 \times 5 = 25 \) possible samples. Each sample can be represented as a pair (e.g., YY, YG, GY, GG), where the first letter represents the outcome of the first selection and the second letter represents the outcome of the second selection.
Calculate the proportion of yellow pods in each sample: For each sample, count the number of yellow pods (Y) and divide by the total number of peas in the sample (2). For example, for the sample YY, the proportion is \( \frac{2}{2} = 1 \), and for the sample YG, the proportion is \( \frac{1}{2} = 0.5 \).
Construct a frequency table: Create a table where each row corresponds to a unique sample, the number of yellow pods in the sample, and the proportion of yellow pods. For example, the table might include rows like (YY, 2, 1), (YG, 1, 0.5), and so on.
Summarize the results: Use the table to analyze the distribution of the proportions of yellow pods across all 25 samples. This can help in understanding the variability and expected outcomes of the hybridization experiment.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hybridization

Hybridization in genetics refers to the process of combining different varieties or species to produce a hybrid. In this context, it involves selecting peas with different pod colors to study the inheritance patterns of traits, such as yellow and green pods. Understanding hybridization is crucial for analyzing genetic variation and predicting the outcomes of breeding experiments.

Sampling with Replacement

Sampling with replacement is a statistical method where each selected item is returned to the population before the next selection. This means that the same pea can be chosen multiple times in the experiment. This technique is important for ensuring that each sample is independent, allowing for a more accurate representation of the population's characteristics.
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Sampling Distribution of Sample Proportion

Proportion

Proportion is a statistical measure that represents the part of a whole, expressed as a fraction or percentage. In this experiment, calculating the proportion of yellow pods in each sample helps to quantify the prevalence of this trait among the selected peas. Understanding proportions is essential for interpreting the results and drawing conclusions about the genetic distribution of traits.
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Difference in Proportions: Hypothesis Tests
Related Practice
Textbook Question

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.


Sampling Distribution of the Sample Standard Deviation For the following, round results to three decimal places.


a. Find the value of the population standard deviation σ.

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Textbook Question

Using the Central Limit Theorem. In Exercises 5–8, assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of 1.2 kg and a standard deviation of 4.9 kg (based on Data Set 13 “Freshman 15” in Appendix B).


a. If 1 male college student is randomly selected, find the probability that he gains at least 2.0 kg during his freshman year..)

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Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.


Redesign of Ejection Seats When women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ACES-II ejection seats were designed for men weighing between 140 lb and 211 lb. Weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 46 lb (based on Data Set 1 “Body Data” in Appendix B).


a. If 1 woman is randomly selected, find the probability that her weight is between 140 lb and 211 lb.

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Textbook Question

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.


Sampling Distribution of the Sample Median


a. Find the value of the population median.

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Textbook Question

Cell Phones and Brain Cancer In a study of 420,095 cell phone users in Denmark, it was found that 135 developed cancer of the brain or nervous system. For those not using cell phones, there is a 0.000340 probability of a person developing cancer of the brain or nervous system. We therefore expect about 143 cases of such cancers in a group of 420,095 randomly selected people.

a. Find the probability of 135 or fewer cases of such cancers in a group of 420,095 people.

b. What do these results suggest about media reports that suggest cell phones cause cancer of the brain or nervous system?

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Textbook Question

Continuity Correction In testing the assumption that the probability of a baby boy is 0.512, a geneticist obtains a random sample of 1000 births and finds that 502 of them are boys. Using the continuity correction, describe the area under the graph of a normal distribution corresponding to the following. (For example, the area corresponding to “the probability of at least 502 boys” is this: the area to the right of 501.5.)


a. The probability of 502 or fewer boys

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