EBK PRACTICE OF STATISTICS IN THE LIFE
EBK PRACTICE OF STATISTICS IN THE LIFE
4th Edition
ISBN: 9781319067496
Author: BALDI
Publisher: VST
Question
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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.

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