EBK PRACTICE OF STATISTICS IN THE LIFE
EBK PRACTICE OF STATISTICS IN THE LIFE
4th Edition
ISBN: 9781319067496
Author: BALDI
Publisher: VST
Question
Book Icon
Chapter 10, Problem 10.8AYK

(a)

To determine

To make: a Venn diagram with information and consider options when labeling the diagram.

(a)

Expert Solution
Check Mark

Answer to Problem 10.8AYK

The Venn diagram was constructed.

Explanation of Solution

Given:

An African study tested 543 children for the sickle-cell anemia trait and for malaria infection. Of those tested, 25% had sickle-cell, 34.6% had malaria and 6.6% had both sickle cell and malaria.

Calculation:

Based on this information a Venn diagram can be constructed. To do this, the probabilities of both malaria and not sickle-cell and sickle-cell and not malaria need to be calculated as follows: P(M and not S)=0.3460.066P(M and not S)=0.28P(S and not M)=0.250.066P(S and not M)=0.184

To calculate the probability that a child has neither sickle cell nor malaria, first the probability a child has either sickle cell or malaria needs to be determined.

  P(S or M)=0.346+0.250.066P(S or M)=0.53

From this, the probability that a child has neither sickle cell nor malaria can be calculated by taking 1 minus this above calculated value.

  P(neither S nor M)=10.53P(neither S nor M)=0.47

These values can now be used to complete the Venn diagram.

  EBK PRACTICE OF STATISTICS IN THE LIFE, Chapter 10, Problem 10.8AYK

Conclusion:

Therefore, the Venn diagram was constructed.

(b)

To determine

To find: the probability that a given child has either malaria or the sickle-cell trait.

(b)

Expert Solution
Check Mark

Answer to Problem 10.8AYK

The probability that a child has either malaria or sickle cell is 0.53.

Explanation of Solution

Calculation:

The probability that a given child has neither malaria nor sickle cell can be seen in the Venn diagram. In addition, to calculate the probability that a child has neither sickle cell nor malaria, first the probability a child has either sickle cell or malaria needs to be determined.

  P(S or M)=0.346+0.250.066P(S or M)=0.53

From this, the probability that a child has neither sickle cell nor malaria can be calculated by taking 1 minus this value.

  P(neither S nor M)=10.53P(neither S nor M)=0.47

The probability that a child has neither sickle cell nor malaria is 0.47.

Also, the probability that a child has either malaria or sickle cell can be determined using the given information. To do this, the probability that a child has sickle cell is added to the probability that a child has malaria. This value is then subtracted from the probability that a child has both.

  P(S or M)=0.346+0.250.066P(S or M)=0.53

Conclusion:

Therefore,the probability that a child has either malaria or sickle cell is 0.53.

(c)

To determine

To find: the probability that a given child has malaria, given that the child has the sickle-cell trait.

(c)

Expert Solution
Check Mark

Answer to Problem 10.8AYK

The probability that a child has malaria given they do not have sickle cell is 0.373.

Explanation of Solution

Calculation:

The probability that a child has malaria given that a child has sickle cell is expressed as follows:

  P(M|SC)=P(M and SC)P(SC)P(M|SC)=P(0.066)P(0.25)P(M|SC)=0.264

The probability that a child has malaria given they have sickle cell is 0.264.

The probability that a child has malaria given that they do not have sickle cell is expressed as follows:

  P(M|notSC)=P(M and not SC)P(not SC)

First, the probability that a child does not have sickle cell needs to be determined by taking 1 minus the probability that a child has sickle cell (0.25).

  P(not SC)=10.25P(not SC)=0.75

Next, substitute the values into the formula given above.

  P(M|notSC)=P(M and not SC)P(not SC)P(M|notSC)=0.280.75P(M|notSC)=0.373

Conclusion:

Therefore,the probability that a child has malaria given they do not have sickle cell is 0.373.

(d)

To determine

To describe: the events sickle-cell trait and malaria independent and might that tell about the relationship between the sickle-cell and malaria.

(d)

Expert Solution
Check Mark

Answer to Problem 10.8AYK

Therefore, the probability of malaria and sickle cell using the multiplication rule gives a probability of 0.0865.As a result, these two events are not independent.

Explanation of Solution

Calculation:

To verify independence, the multiplication of the probabilities of malaria and sickle cell should be equal to the probability of malaria and sickle cell, which is 0.066.

  P(M and SC)=0.066P(M and SC)=P(M)×P(SC)P(M and SC)=0.346×0.25P(M and SC)=0.0865

Calculating the probability of malaria and sickle cell using the multiplication rule gives a probability of 0.0865. This is greater than and does not match the probability of both malaria and sickle cell of 0.066 as given in the example. As a result, these two events are not independent.

Conclusion:

Therefore, the probability of malaria and sickle cell using the multiplication rule gives a probability of 0.0865.As a result, these two events are not independent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
توليد تمرين شامل حول الانحدار الخطي المتعدد بطريقة المربعات الصغرى
The U.S. Postal Service will ship a Priority Mail® Large Flat Rate Box (12" 3 12" 3 5½") any where in the United States for a fixed price, regardless of weight. The weights (ounces) of 20 ran domly chosen boxes are shown below. (a) Make a stem-and-leaf diagram. (b) Make a histogram. (c) Describe the shape of the distribution. Weights 72 86 28 67 64 65 45 86 31 32 39 92 90 91 84 62 80 74 63 86
(a) What is a bimodal histogram? (b) Explain the difference between left-skewed, symmetric, and right-skewed histograms. (c) What is an outlier
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman