/*! 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 76 In Exercise \(5.75,\) a cross be... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 76

In Exercise \(5.75,\) a cross between two peony plants-one with red petals and one with streaky petals-produced offspring plants with red petals \(75 \%\) of the time. Suppose that 100 seeds from this cross were collected and germinated, and \(x\), the number of plants with red petals, was recorded. a. What is the exact probability distribution for \(x ?\) b. Is it appropriate to approximate the distribution in part a using the normal distribution? Explain. c. Use an appropriate method to find the approximate probability that between 70 and 80 (inclusive) offspring plants have red flowers. d. What is the probability that 53 or fewer offspring plants had red flowers? Is this an unusual occurrence? e. If you actually observed 53 of 100 offspring plants with red flowers, and if you were certain that the genetic ratio 3: 1 was correct, what other explanation could you give for this unusual occurrence?

Short Answer

Expert verified
Answer: The approximate probability of observing between 70 and 80 offspring plants with red flowers is 75%.

Step by step solution

01

Identify the type of probability distribution

Since this experiment involves a fixed number of independent trials (germinating 100 seeds) with two possible outcomes (red petals or not red petals) and a constant probability of success (75% chance of red petals), we will use a binomial distribution to model the probability distribution of \(x\).
02

Identify the parameters of the binomial distribution

For a binomial distribution, we need to identify the number of trials (n) and the probability of success (p). In this case, n=100 (100 seeds germinated) and p=0.75 (75% chance of red petals).
03

Answer part a - Calculate the probability distribution

The probability distribution, denoted as \(P(x) = \binom{n}{x} p^x (1-p)^{n-x}\) So, the probability distribution for \(x\) is \(P(x) = \binom{100}{x} (0.75)^x (0.25)^{100-x}\)
04

Answer part b - Determine the appropriateness of a normal approximation

To determine if it is appropriate to use a normal distribution as an approximation, we can apply the rule of thumb that indicates: np >= 10 and n(1-p) >= 10. Here we have np = 100 x 0.75 = 75 and n(1-p) = 100 x 0.25 = 25, both are greater than 10. Thus, it is appropriate to approximate the binomial distribution using a normal distribution.
05

Identify the parameters of the normal distribution

In order to approximate the binomial distribution with a normal distribution, we need the mean (渭) and the standard deviation (蟽). Mean (渭) = np = 100 x 0.75 = 75 Variance = np(1-p) = 100 x 0.75 x 0.25 = 18.75 Standard Deviation (蟽) = 鈭歏ariance = 鈭�18.75 鈮� 4.33
06

Answer part c - Calculate the probability of between 70 and 80 red petals

We will use the normal distribution to approximate this probability. First, we normalize the values of 70 and 80 using the z-score formula: \(z_1 = \frac{70-75}{4.33} \approx-1.15\) \(z_2 = \frac{80-75}{4.33} \approx1.15\) Now, using a z-table or a calculator to find the area between these z-scores, we get P(\(-1.15 \le z \le 1.15) \approx 0.75\). So, the approximate probability that between 70 and 80 offspring plants have red flowers is 75%.
07

Answer part d - Calculate the probability of 53 or fewer red petals

First, we normalize the value of 53 using the z-score formula: \(z_3 = \frac{53-75}{4.33} \approx -5.08\) Using a z-table or calculator to find the area to the left of this z-score, we get P(\(z \le -5.08) \approx 0\) (the probability is too close to 0 to register using the normal approximation). This indicates that it is an extremely unusual occurrence to have 53 or fewer offspring plants with red flowers.
08

Answer part e - Possible explanation for unusual occurrence

If you observed 53 of 100 offspring plants with red flowers and were certain that the genetic ratio 3:1 was correct, a possible explanation for this unusual occurrence could be that: 1. The sample of seeds may not be representative of the overall cross-breed population; 2. There might be some errors or biases in the seed selection or germination process; 3. There could be some external environmental factors affecting the germination of seeds or the expression of red petals in offspring plants.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Binomial Distribution
When dealing with probability, the binomial distribution is a central concept that allows us to handle situations where we have a series of trials with two distinct outcomes typically labeled as 'success' and 'failure'. In the given exercise, the success is defined as a seed germinating into a plant with red petals.

The binomial distribution is specifically characterized by three criteria: a fixed number of trials (in the exercise, 100 seeds), each trial is independent of the others, and the probability of success is the same for each trial (here, a 75% chance for red petals). The mathematical representation of the probability distribution is given by the formula: \[ P(x) = \binom{n}{x} p^x (1-p)^{n-x} \],where \( n \) represents the total number of trials, \( x \) is the number of successes (red-petaled plants), \( p \) is the probability of success, and \( 1-p \) is the probability of failure. This allows students to calculate the likelihood of obtaining any given number of successes in the experiment.
Normal Approximation
The normal approximation is a handy method for simplifying the computation of probabilities when dealing with large sample sizes. It's especially useful when the binomial distribution has a high number of trials (\( n \)), making the calculations cumbersome. However, it's not suitable in every case. It's commonly applicable when both the minimum expected successes (\( np \)) and failures (\( n(1-p) \)) are greater than 10.

In our case, both these numbers are quite large (75 and 25, respectively), therefore, the normal approximation to the binomial distribution is appropriate. The normal distribution is a continuous probability distribution that is symmetrical around its mean, resembling a bell-shaped curve. It simplifies our computations for binomial distributions with higher trials as we can use z-scores to make it easier to understand and find the probabilities for different ranges of successes. This approximation streamlines the process, as one can reference standard z-tables or use software to quickly find probabilities.
Z-Score Calculation
A z-score is a numerical measurement used in statistics to describe a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a z-score is 0, it indicates that the data point's score is identical to the mean score.

A z-score is calculated via the formula: \[ z = \frac{(x - \mu)}{\sigma} \],where \( x \) is the value being investigated, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation of the distribution. This conversion to a z-score allows for the comparison between different data points within the distribution and aids in understanding the probability of a score occurring within a normal distribution. In our exercise context, we calculated z-scores to understand the likelihood of having 70 to 80 plants with red petals and determine whether or not 53 red-petaled plants is a common occurrence. This standardization is a powerful statistical tool that helps in making informed conclusions about the observed data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article in Time magazine \(^{7}\) (Exercise 6.52 ) also reported that \(80 \%\) of men and \(62 \%\) of women put in more than 40 hours a week on the job. Assume that these percentages are correct for all Americans, and that a random sample of 50 working women is selected. a. What is the average number of women who put in more than 40 hours a week on the job? b. What is the standard deviation for the number of women who put in more than 40 hours a week on the job? c. Suppose that in our sample of 50 working women, there are 25 who work more than 40 hours a week. Would you consider this to be an unusual occurrence? Explain.

The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company's drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c. What proportion will have to be replaced after more than 90 hours of use?

Two of the biggest soft drink rivals, Pepsi and Coke, are very concerned about their market shares. The pie chart that follows claims that PepsiCo's share of the beverage market is \(25 \% .^{6}\) Assume that this proportion will be close to the probability that a person selected at random indicates a preference for a Pepsi product when choosing a soft drink. A group of \(n=500\) consumers is selected and the number preferring a Pepsi product is recorded. Use the normal curve to approximate the following binomial probabilities. a. Exactly 150 consumers prefer a Pepsi product. b. Between 120 and 150 consumers (inclusive) prefer a Pepsi product. c. Fewer than 150 consumers prefer a Pepsi product. d. Would it be unusual to find that 232 of the 500 consumers preferred a Pepsi product? If this were to occur, what conclusions would you draw?

Consider a binomial random varible with \(n=25\) and \(p=.6 .\) Fill in the blanks below to find some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate \(n p=\) _____ and \(n q=\) _____ b. Are \(n p\) and \(n q\) both greater than \(5 ?\) Yes ____ No ____ c. If the answer to part \(b\) is yes, calculate \(\mu=n p=\) ______ and \(\sigma=\sqrt{n p q}=\) ______ d. To find the probability of more than 9 successes, what values of \(x\) should be included? \(x=\) ________ e. To include the entire block of probability for the first value of \(x=\) ______, start at _______. f. Calculate \(z=\frac{x \pm .5-n p}{\sqrt{n p q}}=\) _______. g. Calculate \(P(x>9) \approx P(z>\)______) \(=1-\) _____ \(=\) ____.

Calculate the area under the standard normal curve between these values: a. \(z=-2.0\) and \(z=2.0\) b. \(z=-2.3\) and -1.5
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.