LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)
LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)
8th Edition
ISBN: 9781464133404
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 5.2, Problem 43UYK

(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
Check Mark

Answer to Problem 43UYK

Solution: The probability is 0.9953.

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(1p)n=0.5(10.5)200=0.25200=0.03536

Hence, the average value and standard deviation are 0.5 and 0.03536, respectively. Now, calculate the z-score,

z=(xμ)s

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

z1=(xμ)s=(0.40.5)0.0356=2.83

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

z2=(xμ)s=(0.60.5)0.03536=2.83

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(2.83) to calculate the left-tailed area of z1. The screenshot is shown below:

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access), Chapter 5.2, Problem 43UYK , additional homework tip  1

Use the formula =NORMSDIST(2.83) to calculate the left tailed area of z2. The screenshot is shown below:

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access), Chapter 5.2, Problem 43UYK , additional homework tip  2

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

(z1<z<z2)=0.9976730.002327=0.9953

Hence, the probability is 0.9953.

(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
Check Mark

Answer to Problem 43UYK

Solution: The probability is 0.8426.

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(1p)n=0.5(10.5)200=0.25200=0.03536

Hence, the average value and standard deviation are 0.5 and 0.03536 respectively. Now, calculate the z-score,

z=(xμ)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=(xμ)s=(0.450.5)0.03536=1.414

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

z2=(xμ)s=(0.550.5)0.03536=1.414

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula =NORMSDIST(1.414) to calculate the left tailed area of z1. The screenshot is shown below:

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access), Chapter 5.2, Problem 43UYK , additional homework tip  3

Use the formula =NORMSDIST(1.414) to calculate the left tailed area of z2. The screenshot is shown below:

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access), Chapter 5.2, Problem 43UYK , additional homework tip  4

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

(z1<z<z2)=0.921320.07868=0.8426

Hence, the probability is 0.8426

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Chapter 5 Solutions

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.1 - Prob. 31ECh. 5.1 - Prob. 32ECh. 5.1 - Prob. 33ECh. 5.1 - Prob. 34ECh. 5.2 - Prob. 35UYKCh. 5.2 - Prob. 36UYKCh. 5.2 - Prob. 37UYKCh. 5.2 - Prob. 38UYKCh. 5.2 - Prob. 39UYKCh. 5.2 - Prob. 40UYKCh. 5.2 - Prob. 41UYKCh. 5.2 - Prob. 42UYKCh. 5.2 - Prob. 43UYKCh. 5.2 - Prob. 44UYKCh. 5.2 - Prob. 45UYKCh. 5.2 - Prob. 46ECh. 5.2 - Prob. 47ECh. 5.2 - Prob. 48ECh. 5.2 - Prob. 49ECh. 5.2 - Prob. 50ECh. 5.2 - Prob. 51ECh. 5.2 - Prob. 52ECh. 5.2 - Prob. 53ECh. 5.2 - Prob. 54ECh. 5.2 - Prob. 55ECh. 5.2 - Prob. 56ECh. 5.2 - Prob. 57ECh. 5.2 - Prob. 58ECh. 5.2 - Prob. 59ECh. 5.2 - Prob. 60ECh. 5.2 - Prob. 61ECh. 5.2 - Prob. 62ECh. 5.2 - Prob. 63ECh. 5.2 - Prob. 64ECh. 5.2 - Prob. 65ECh. 5.2 - Prob. 66ECh. 5.2 - Prob. 67ECh. 5.2 - Prob. 68ECh. 5.2 - Prob. 69ECh. 5.2 - Prob. 70ECh. 5.2 - Prob. 71ECh. 5.2 - Prob. 72ECh. 5.2 - Prob. 73ECh. 5.2 - Prob. 74ECh. 5.2 - Prob. 75ECh. 5 - Prob. 76ECh. 5 - Prob. 77ECh. 5 - Prob. 78ECh. 5 - Prob. 79ECh. 5 - Prob. 80ECh. 5 - Prob. 81ECh. 5 - Prob. 82ECh. 5 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90E
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