Theprobability that exactly 3 blood type O of the given data.

Question

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Answer 1

Question

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

p[X≤3]=p[X=0]+p[X=1]+p[X=2]+p[X=3]p[X≤3]=0.0425

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Answer 2

Question

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

p[X≤3]=p[X=0]+p[X=1]+p[X=2]+p[X=3]p[X≤3]=0.0425

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Answer 3

Question

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

p[X≤3]=p[X=0]+p[X=1]+p[X=2]+p[X=3]p[X≤3]=0.0425

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Answer 4

Question

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

p[X≤3]=p[X=0]+p[X=1]+p[X=2]+p[X=3]p[X≤3]=0.0425

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Answer 5

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

p[X≤3]=p[X=0]+p[X=1]+p[X=2]+p[X=3]p[X≤3]=0.0425

Answer 6

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

Answer 7

Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly 3 blood type O of the given data.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318

p[X≤3]=p[X=0]+p[X=1]+p[X=2]+p[X=3]p[X≤3]=0.0425

Answer 13

(b)

To determine

To find:

The probability that 3 or fewer is blood type O of the given data.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

m=15p=0.45q=0.55

Concept used:

Formula

The binomial distribution

p[X=x]=(nx)pxqn−x

Calculation:

Using by the binomial distribution

p[X=x]=(nx)pxqn−x

p[X=0]=(150)(0.45)0(0.55)15−0p[X=0]=0.0001p[X=1]=(151)(0.45)1(0.55)14p[X=1]=0.0016p[X=2]=(152)(0.45)2(0.55)13p[X=2]=0.0090p[X=3]=(153)(0.45)3(0.55)12p[X=3]=0.0318