The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

Question

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Answer 1

Question

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below:

Use the formula =NORMSDIST(1.23) to calculate the left tailed area of z2. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.89065−0.10935=0.7813

Hence, the probability is 0.7813

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Answer 2

Question

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below:

Use the formula =NORMSDIST(1.23) to calculate the left tailed area of z2. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.89065−0.10935=0.7813

Hence, the probability is 0.7813

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Answer 3

Question

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below:

Use the formula =NORMSDIST(1.23) to calculate the left tailed area of z2. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.89065−0.10935=0.7813

Hence, the probability is 0.7813

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below:

Use the formula =NORMSDIST(1.23) to calculate the left tailed area of z2. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.89065−0.10935=0.7813

Hence, the probability is 0.7813

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below:

Use the formula =NORMSDIST(1.23) to calculate the left tailed area of z2. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.89065−0.10935=0.7813

Hence, the probability is 0.7813

Answer 6

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

Answer 7

Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

Answer 8

(a)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.4 and the upper bound si x2=0.6 with mean 0.5. Calculate the z- score for lower bound x1=0.4 as:

z1=(x−μ)s=(0.4−0.5)0.0408=−2.45

Calculate the z- score for the upper bound x2=0.6 as:

z2=(x−μ)s=(0.6−0.5)0.0408=2.45

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−2.45) to calculate the left-tailed area of z1. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 1

Use the formula =NORMSDIST(−2.45) to calculate the left tailed area of z2. The screenshot is shown below:

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access), Chapter 5.3, Problem 53UYK , additional homework tip 2

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.992857−0.007143=0.9857

Hence, the probability is 0.9857.

Answer 13

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below:

Use the formula =NORMSDIST(1.23) to calculate the left tailed area of z2. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

(z1<z<z2)=0.89065−0.10935=0.7813

Hence, the probability is 0.7813

Answer 14

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

Answer 15

(b)

Expert Solution

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

p=12=0.5

The average of the sample proportion is calculated as:

μp^=p=0.5

The standard deviation of the sample proportion is calculated as:

σp^=p(1−p)n=0.5(1−0.5)150=0.25150=0.0408

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

z=(x−μ)s

The lower bound is x1=0.45 and the upper bound is x2=0.55 with mean 0.5. Calculate the z- score for lower bound x1=0.45 as:

z1=(x−μ)s=(0.45−0.5)0.0408=−1.23

Calculate the z- score for the upper bound x2=0.55 as:

z2=(x−μ)s=(0.55−0.5)0.0408=1.23

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(−1.23) to calculate the left tailed area of z1. The screenshot is shown below: