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Pea plants contain two genes for seed color, each of which

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 3SE Chapter 4

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 3SE

Pea plants contain two genes for seed color, each of which may be Y (for yellow seeds) or G (for green seeds). Plants that contain one of each type of gene are called heterozygous. According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability 0.75 of having yellow seeds and probability 0.25 of having green seeds.

a. Out of 10 offspring of heterozygous plants, what is the probability that exactly 3 have green seeds?

b. Out of 10 offspring of heterozygous plants, what is the probability that more than 2 have green seeds?

c. Out of 100 offspring of heterozygous plants, what is the probability that more than 30 have green seeds?

d. Out of 100 offspring of heterozygous plants, what is the probability that between 30 and 35 inclusive have green seeds?

e. Out of 100 offspring of heterozygous plants, what is the probability that fewer than 80 have yellow seeds?

Step-by-Step Solution:
Step 1 of 3

Solution 3SE

Step1 of 6:

Let us consider a pie plant it contains two genes for seeds colour.

Y = yellow seeds and

G = green seeds.

Here each type of gene is called as heterozygous.

Mendelian theory of genetics is stats that:

If two heterozygous plants are crossed, Then

P(Each of their offspring of having yellow seeds) = 0.75 and

P(Each of their offspring of having green seeds) = 0.25.

Let X follows binomial distribution with parameters “n and p ”.

Then the probability mass function of binomial distribution is given by:

, x = 0,1,2,...,n. 

Where,

n = sample size

x = random variable

p = probability of success

q = 1 - p (probability of failure)

Here our goal is:

a).We need to find the probability that exactly 3 have green seeds, when Out of 10 offspring of heterozygous plants.

b).We need to find the probability that more than 2 have green seeds, when Out of 10 offspring of heterozygous plants.

c).We need to find the probability that more than 30 have green seeds, when Out of 100 offspring of heterozygous plants.

d).We need to find the probability that between 30 and 35 inclusive have green seeds, when Out of 100 offspring of heterozygous plants.

e).We need to find the probability that fewer than 80 have yellow seeds, when Out of 100 offspring of heterozygous plants.

Step2 of 6:

a).

We have n = 10 and P(Each of their offspring of having green seeds) = 0.25 that is p = 0.25 and

x = 3

Now,

XB(n, p)

       XB(10, 0.25)

Consider the pmf:

, x = 0,1,2,...,n.  

Substitute n, p and x value in above equation then

 

                                 

                                             

                                             

                                             

                                             

                                              0.250

Hence, P(Exactly 3 have green seeds) = 0.250.

Step3 of 6:

b).

We have n = 10 and P(Each of their offspring of having green seeds) = 0.25 that is p = 0.25 and

x > 2.

Now,

XB(n, p)

       XB(10, 0.25)

Consider the pmf:

, x = 0,1,2,...,n.  

Substitute n, p and x value in above equation then

 

 

                   

Consider, can be obtained from Excel by using function “=Binomdist(X,n,p,FALSE)” 

...

X

Function

P(X=x)

0

Binorm(0,10,0.25,FALSE)”

0.05631351471

1

Binorm(1,10,0.25,FALSE)”

Step 2 of 3

Chapter 4, Problem 3SE is Solved
Step 3 of 3

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

Statistics for Engineers and Scientists was written by and is associated to the ISBN: 9780073401331. This textbook survival guide was created for the textbook: Statistics for Engineers and Scientists , edition: 4. This full solution covers the following key subjects: Seeds, plants, heterozygous, Probability, offspring. This expansive textbook survival guide covers 153 chapters, and 2440 solutions. The full step-by-step solution to problem: 3SE from chapter: 4 was answered by , our top Statistics solution expert on 06/28/17, 11:15AM. Since the solution to 3SE from 4 chapter was answered, more than 555 students have viewed the full step-by-step answer. The answer to “Pea plants contain two genes for seed color, each of which may be Y (for yellow seeds) or G (for green seeds). Plants that contain one of each type of gene are called heterozygous. According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability 0.75 of having yellow seeds and probability 0.25 of having green seeds.a. Out of 10 offspring of heterozygous plants, what is the probability that exactly 3 have green seeds?________________b. Out of 10 offspring of heterozygous plants, what is the probability that more than 2 have green seeds?________________c. Out of 100 offspring of heterozygous plants, what is the probability that more than 30 have green seeds?________________d. Out of 100 offspring of heterozygous plants, what is the probability that between 30 and 35 inclusive have green seeds?________________e. Out of 100 offspring of heterozygous plants, what is the probability that fewer than 80 have yellow seeds?” is broken down into a number of easy to follow steps, and 157 words.

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