Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
bartleby

Videos

Question
Book Icon
Chapter 5.3, Problem 78E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

(a)

Expert Solution
Check Mark

Answer to Problem 78E

Solution: The mean and standard deviation of the binomial random variable X is 996 and 13.012 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

μx=np

In binomial distribution, standard deviation can be calculated by this formula,

σx=npq

Where, n is the number of trials and p is the probability of success in a trial. For x=0,1,2,3...1200, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

μX=np=1200×0.83=996

The standard deviation is calculated as:

σX=np(1p)=1200(0.83)(10.83)=13.012

Hence, the average value and standard deviation are 996 and 13.012 respectively.

(b)

To determine

To find: The probability.

(b)

Expert Solution
Check Mark

Answer to Problem 78E

Solution: The probability is P(X800)=1_.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean μ=np and standard deviation σ=np(1p) if the condition np(1p)10 holds true, where n is the number of trials and p is the probability of success in a trial.

np(1p)=1200×0.83×(10.83)=169.32>10

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

μ=np=1200×0.83=996

And:

σ=np(1p)=1200(0.83)(10.83)=13.012

The probability that the value taken by X is greater than or equal to 800 is the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value X=800 is calculated as:

Z=Xμσ=80099613.012=15.06

The area left to the particular Z-score can be obtained using Excel by using the command '=NORMSDIST()'. Specify the value of z as 15.06.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip  1

The area left to the Z-score, 15.06, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

10.0000=1

Hence, the probability is 1.

(c)

To determine

To find: The probability.

(c)

Expert Solution
Check Mark

Answer to Problem 78E

Solution: P(X>1000)=0.3676_.

Explanation of Solution

Calculation: For the probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cells and specify the value of number=1000,trials = 1200, probability = 0.83,and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip  2

P(X1000)=0.6324.

Therefore, the required probability is calculated as:

P(X>1000)=1P(X1000)=10.6324=0.3676

Hence, the probability is 0.3676.

(d)

To determine

To find: The probability.

(d)

Expert Solution
Check Mark

Answer to Problem 78E

Solution: P(X>1000)=0.0001_.

Explanation of Solution

Calculation: The probability that more than 1000 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command '=BINOMDIST()' in one of the cell and specify the value of number=1000,trials = 1150, probability = 0.83,and cumulative as 'TRUE'. Press enter to obtain the probability that the value of the random variable X is less than or equal to 1000.

Introduction to the Practice of Statistics, Chapter 5.3, Problem 78E , additional homework tip  3

P(X1000)=0.9999

Therefore, the required probability is calculated as:

P(X>1000)=1P(X1000)=10.9999=0.0001

Hence, the probability is 0.0001.

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
0|0|0|0 - Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were performed on Xt to obtain Yt? seasonal difference of order 2 simple difference of order 5 seasonal difference of order 1 seasonal difference of order 5 simple difference of order 2
Calculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.
Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADO

Chapter 5 Solutions

Introduction to the Practice of Statistics

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. 5.2 - Prob. 17UYKCh. 5.2 - Prob. 18UYKCh. 5.2 - Prob. 19UYKCh. 5.2 - Prob. 20UYKCh. 5.2 - Prob. 21UYKCh. 5.2 - Prob. 22UYKCh. 5.2 - Prob. 23UYKCh. 5.2 - Prob. 24UYKCh. 5.2 - Prob. 25ECh. 5.2 - Prob. 26ECh. 5.2 - Prob. 27ECh. 5.2 - Prob. 28ECh. 5.2 - Prob. 29ECh. 5.2 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.3 - Prob. 43UYKCh. 5.3 - Prob. 44UYKCh. 5.3 - Prob. 45UYKCh. 5.3 - Prob. 46UYKCh. 5.3 - Prob. 47UYKCh. 5.3 - Prob. 48UYKCh. 5.3 - Prob. 49UYKCh. 5.3 - Prob. 50UYKCh. 5.3 - Prob. 51UYKCh. 5.3 - Prob. 52UYKCh. 5.3 - Prob. 53UYKCh. 5.3 - Prob. 54UYKCh. 5.3 - Prob. 55UYKCh. 5.3 - Prob. 56UYKCh. 5.3 - Prob. 57ECh. 5.3 - Prob. 58ECh. 5.3 - Prob. 59ECh. 5.3 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90ECh. 5 - Prob. 91ECh. 5 - Prob. 92ECh. 5 - Prob. 93ECh. 5 - Prob. 94ECh. 5 - Prob. 95ECh. 5 - Prob. 96ECh. 5 - Prob. 97ECh. 5 - Prob. 98ECh. 5 - Prob. 99ECh. 5 - Prob. 100ECh. 5 - Prob. 101ECh. 5 - Prob. 102E
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
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
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License