Rounded to 4 decimal, what is the standard deviation of a binomila distribution that uses 216 trials with a success probability of 0.17?

Rounded to 4 decimal, what is the standard deviation of a binomila distribution that uses 23 trials with a success probability of 0.85?

Rounded to 4 decimal, what is the standard deviation of a binomila distribution that uses 31 trials with a success probability of 0.19?

(A) Discuss the probability of landing on heads if you flipped a coin 10 times?

(B) What is the probability the coin will land on heads on each of the 10 coin flips?

(C) Apply this same binomial experiment to a different real-world situation. Describe a situation involving probability?

please explain each and show work. showing the steps to the answer would be great..

3) The momentum eigenfunction for a particle moving in one dimension is фр--h-1/2eipz/n The energy eigenfunction for a particle in a 1D box of length L is u()- is expanded in terms of фе(x), the expansion coefficient may be interpreted as the momentum probability amplitude; its square gives the probability density for momentum. Determine the momentum probability density for u(x)

Write a class to implement HeadTailListInterface. Instead of using an array, use a List object as your instance data variable. (List (Links to an external site.) from the Java standard library- not ListInterface!). Instantiate the List object to type ArrayList.

Inside the methods of this class, invoke methods on the List object to accomplish the task. Note: some methods might look very simple... this does not mean they are wrong!

There is one difference in how this class will work compared to the other: in this extra credit class, you do not have control over the capacity, so you should not print the capacity in display and the capacity does not have to be exactly doubled in the two add methods.

For full credit:

• Pay close attention to what should happen in "failed" conditions as described by the HeadTailListInterface compared to List!
• Make sure your ListHeadTailList behaves as described in HeadTailListInterface.
• Use the methods of the List interface/ArrayList class when possible instead of re-writing the code yourself.

The class header and instance data will be:

public class ListHeadTailList<T> implements HeadTailListInterface<T>

List<T> list; // initialize to type ArrayList<T> in the ListHeadTailList constructor

------------------------------------------------------------------------------------------------------------------------------------------------

/**
* An interface for a list. Entries in a list have positions that begin with 0.
* Entries can only be removed or added to the beginning and end of the list.
* Entries can be accessed from any position.
*
* @author Jessica Masters
*/
public interface HeadTailListInterface<T> {
  
   /**
   * Adds a new entry to the beginning of the list.
   * Entries currently in the list are shifted down.
   * The list size is increased by 1.
   *
   * @param newEntry The object to be added as a new entry.
   */
   public void addFront(T newEntry);
  
   /**
   * Adds a new entry to the end of the list.
   * Entries currently in the list are unaffected.
   * The list size is increased by 1.
   *
   * @param newEntry The object to be added as a new entry.
   */
   public void addBack(T newEntry);

   /**
   * Removes an entry from the beginning of the list.
   * Entries currently in the list are shifted up.
   * The list size is decreased by 1.
   *
   * @return A reference to the removed entry or null if the list is empty.
   */
   public T removeFront();
  
   /**
   * Removes an entry from the end of the list.
   * Entries currently in the list are unaffected.
   * The list size is decreased by 1.
   *
   * @return A reference to the removed entry or null if the list is empty.
   */
   public T removeBack();

  
   /** Removes all entries from this list. */
   public void clear();

  
   /**
   * Retrieves the entry at a given position in this list.
   *
   * @param givenPosition An integer that indicates the position of the desired entry.
   * @return A reference to the indicated entry or null if the index is out of bounds.
   */
   public T getEntry(int givenPosition);
  
   /**
   * Displays the contents of the list to the console, in order.
   */
   public void display();
  
   /**
   * Determines the position in the list of a given entry.
   * If the entry appears more than once, the first index is returned.
   *
   * @param anEntry the object to search for in the list.
   * @return the first position the entry that was found or -1 if the object is not found.
   */
   public int indexOf(T anEntry);
  
   /**
   * Determines the position in the list of a given entry.
   * If the entry appears more than once, the last index is returned.
   *
   * @param anEntry the object to search for in the list.
   * @return the last position the entry that was found or -1 if the object is not found.
   */
   public int lastIndexOf(T anEntry);
  
   /**
   * Determines whether an entry is in the list.
   *
   * @param anEntry the object to search for in the list.
   * @return true if the list contains the entry, false otherwise
   */
   public boolean contains(T anEntry);

   /**
   * Gets the length of this list.
   *
   * @return The integer number of entries currently in the list.
   */
   public int size();

   /**
   * Checks whether this list is empty.
   *
   * @return True if the list is empty, or false if the list contains one or more elements.
   */
   public boolean isEmpty();
}

------------------------------------------------------------------------------------------------------------------------

********TESTING ISEMPTY AND EMPTY DISPLAY
Empty is true: true

Should display:
0 elements; capacity = 10
0 elements; capacity N/A        


********TESTING ADD TO FRONT
Should display:
1 elements; capacity = 10       [2]
1 elements; capacity N/A        [2]

Should display:
3 elements; capacity = 10       [3, 4, 2]
3 elements; capacity N/A        [3, 4, 2]

Empty is false: false


********TESTING CLEAR
Should display:
0 elements; capacity = 10
0 elements; capacity N/A        

********TESTING ADD TO BACK
Should display:
1 elements; capacity = 10       [7]
1 elements; capacity N/A        [7]

Should display:
3 elements; capacity = 10       [7, 10, 5]
3 elements; capacity N/A        [7, 10, 5]


********TESTING CONTAINS
Contains 5 true:  true
Contains 7 true:  true
Contains 4 false: false


********TESTING ADD WITH EXPANSION
Should display:
32 elements; capacity = 40      [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]
32 elements; capacity N/A       [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]


********TESTING INDEX OF
Index of 0  is  0: 0
Index of 31 is 31: 31
Index of -5 is -1: -1
Index of 32 is -1: -1
Index of 3  is  0: 0
Index of 5  is  6: 6


********TESTING LAST INDEX OF
Last index of 0  is  1:  1
Last index of 31 is 32: 32
Last index of -5 is -1: -1
Last index of 35 is -1: -1
Last index of 3  is  4:  4
Last index of 5  is  33: 33


********TESTING SIZE
Size is 34: 34


********TESTING GET ENTRY
Element in position 15 is 14: 14
Element in position  0 is  3: 3
Element in position 39 is null: null
Element in position -1 is null: null


********TESTING REMOVES
Remove front element 3: 3
Remove back element  5 :5
Remove front element 0: 0
Remove back element 31: 31

Should display:
30 elements; capacity = 40      [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30]
30 elements; capacity N/A       [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30]

Remove element null: null
Remove element null: null

Remove element 1: 1
Should display:
0 elements; capacity = 40
0 elements; capacity N/A        

Remove element 1: 1
Should display:
0 elements; capacity = 40
0 elements; capacity N/A        

Remove element 1: 1
Should display:
0 elements; capacity = 40
0 elements; capacity N/A        

Remove element 1: 1
Should display:
0 elements; capacity = 40
0 elements; capacity N/A        



********TESTING MIX OF ADDS AND REMOVES
Should display:
7 elements; capacity = 40       [5, 4, 3, 2, 3, 8, 9]
7 elements; capacity N/A        [5, 4, 3, 2, 3, 8, 9]

Should display:
5 elements; capacity = 40       [4, 3, 2, 3, 8]
5 elements; capacity N/A        [4, 3, 2, 3, 8]

********TESTING WITH STRINGS
Should display:
5 elements; capacity = 5        [You, did, it!, Nice, job!]
5 elements; capacity = 5        [You, did, it!, Nice, job!]

Contains "Nice" is true: true
Contains "You"  is true: true
Contains "you"  is false: false

Index of "it!" is 2: 2
Last index of "it!" is 2: 2

1. Suppose x is a random variable best described by a uniform probability distribution with   and   Find  

2. Suppose x is a random variable best described by a uniform probability distribution with   and   Find  

3. Suppose x is a random variable best described by a uniform probability distribution with   and   Find  

4. Suppose x is a random variable best described by a uniform probability distribution with   and   Find  

5. Suppose x is a random variable best described by a uniform probability distribution with   and   Find  

6. Suppose x is a random variable best described by a uniform probability distribution with   and   Find  

7. Suppose x is a uniform random variable with c = 10 and d = 90. Find P(13 < x < 85).
Round to the nearest hundredth when necessary.

8. Suppose x is a uniform random variable with c = 10 and d = 50. Find P(13 < x < 45).
Round to the nearest hundredth when necessary.

9. Suppose x is a uniform random variable with c = 20 and d = 70. Find P(23 < x < 65).
Round to the nearest hundredth when necessary.

10. High temperatures in a certain city for the month of August follow a uniform distribution over the interval   to   Find the temperature which is exceeded by the high temperatures on 90% of the days in August.

11. Suppose a uniform random variable can be used to describe the outcome of an experiment with outcomes ranging from 20 to 70. What is the probability that this experiment results in an outcome less than 30? Round to the nearest hundredth when necessary

12. Suppose x is a random variable best described by a uniform probability distribution with   and   Find the value of a that makes the following probability statement true:  

13. The diameters of ball bearings produced in a manufacturing process can be described using a uniform distribution over the interval 6.5 to 8.5 millimeters. What is the probability that a randomly selected ball bearing has a diameter greater than 7.4 millimeters?

Part 1: Combinations and Permutations: Winning the Lottery

To win the Powerball jackpot you need to choose the correct five numbers from the integers 1 - 69 as well as pick the correct Powerball which is one number picked from the integers 1 - 26. The order in which you pick the numbers is not relevant. You just need to pick the correct five numbers in any order and the correct Powerball.

Because there is only one correct set of five numbers and one correct Powerball, the probability of winning the jackpot would be calculated as:

  

To calculate the “# of ways of choosing the numbers” we use combinations.

The expression for combinations is nCk, where n is the number of items available to be chosen from and k is the number of items chosen.

For the portion of Powerball where 5 numbers are chosen from 1 – 69, n=69 and k=5. The number of ways to choose five numbers from the integers 1 - 69 is calculated as:

  

The symbol ! is called “factorial.” The Factorial of a Natural Number is the product of the number and all natural numbers below it.

For instance, 4! = 4321 = 24.

So 69C5 can be simplified as:

69C5 =
69! 5!(69−5)!
=
69∙68∙67∙66∙65∙64! 5!64!
=
69∙68∙67∙66∙65 5!
=
69∙68∙67∙66∙65 5∙4∙3∙2∙1
= 11,238,513


This means there are 11,238,513 different ways to choose 5 numbers from the integers 1 – 69 when the order of the numbers does not matter. If there was a lottery game based on choosing 5 numbers from 1 – 69, the probability of winning would be P=1/11,238,513.

Since this is the Powerball we also need to choose one number between the integers 1 - 26. We know there are 26 ways to choose one number out of 26 numbers but we can use combinations again if we wanted to. That calculation would be:

26C1 =
26! 1!(26−1)!
=
26∙25! 1!25!
= 26

3

Since choosing the five numbers is independent of choosing the Powerball, the total number of ways of choosing the numbers is:



Therefore, the probability of winning the Powerball jackpot is P =
1 292,201,338 .


Part 1 Exercises: Use the above example as a guide to answer the following questions. You must explain how the probabilities are calculated to earn full credit. Express your answers as fractions and use commas in large numbers.

1. Use the link to find information about the Mega Millions game: http://www.flalottery.com/ Once on the website, click on “Mega Millions”, and then on “How to Play and How to Win” to get the information needed to calculate the probability of winning the Mega Millions jackpot. Find the probability of winning the Mega Millions jackpot using the same manner as the powerball example.

2. If a lottery game was based on choosing 5 numbers from the integers 1 – 52, what is the probability of choosing the numbers for the winning jackpot?

3. If a lottery game was based on choosing 6 numbers from the integers 1 – 49 with another number chosen from the integers 1 – 22, what is the probability of choosing the numbers for the winning jackpot?




Part 2: Descriptive Statistics

Example: A call center monitored the length of time needed to resolve a customer’s service issue. For 8 customers, the time in minutes to resolve each issue is listed in the following data set. 18, 24, 45, 21, 22, 31, 16, 21

The manager of the call center wants to know the mean, median, mode, and standard deviation of the data set.

The mean is calculated with the formula: ?̅ =
Σ? ?

The sample mean is ?̅ . x = data value n = sample size = number of data values

There are 8 data values, so n=8.

4

The mean is: ?̅ =
18+24+45+21+22+31+16+21 8
=
198 8
= 24.75 Because the data values are integers, the mean would be rounded to one more decimal place, giving a mean of ?̅ = 24.8. If the mean is used for further calculations (such as for standard deviation, shown later in this document), it should not be rounded during those calculations. The rounding is done after the calculations (the standard deviation will be rounded).


The median of a data set, denoted as ? ̃, is the “middle” value of the data set. The first step in finding the median is sorting the data set in ascending order (from smallest to largest), as shown below.

16, 18, 21, 21, 22, 24, 31, 45

If there is an odd number of data values, ? ̃ is the actual middle value of the data set. Since this data set consists of an even number of data values the median is the average of the two middle values. There are 8 data values (n=8) and n/2 = 4. Counting in 4 values from the left gives a data value of 21. Counting in 4 values from the right gives a data value of 22.

The median is: ? ̃ =
21+22 2
= 21.5

The mode of a data set is the value that occurs most often. For this data set the mode is 21 because it occurs twice. It is possible for a data set to have more than one mode. If the data value 16 occurred twice in the data set, the two modes would be 21 and 16. It is also possible for a data set to not have a mode. If every data value occurs only one time, the data set would not have a mode.

The standard deviation of a data set (a sample), denoted as s, describes how much the data set “varies” or how it is “dispersed” around the mean. The formula for standard deviation is shown below. The formula for standard deviation is: ? = √Σ(?−?̅)2 ?−1


To calculate standard deviation, it is sometimes convenient to use a table, as shown below, to keep track of the value for (? − ?)2. The first step is to find the difference between the data value and the mean, which was found above, ?̅ = 24.75 . This is shown in the second column of the table. x (data value) (? − ?̅) (? − ?̅)2 18 18-24.75= -6.75 24 24-24.75= -0.75 45 45-24.75= 20.25 21 21-24.75= -3.75 22 22-24.75= -2.75 31 31-24.75= 6.25 16 16-24.75= -8.75 21 21-24.75 = -3.75
5

The next step is to square the difference between the data value and the mean. This can be done immediately after finding the difference when using a calculator by just pressing the squaring button. This is shown in the third column of the table. Note that these values are not rounded.

x (data value) (? − ?̅) (? − ?̅)2 18 18-24.75= -6.75 (-6.75)2=45.5625 24 24-24.75= -0.75 (-0.75)2=0.5625 45 45-24.75= 20.25 (20.25)2=410.0625 21 21-24.75= -3.75 (-3.75)2=14.0625 22 22-24.75= -2.75 (-2.75)2=7.5625 31 31-24.75= 6.25 (6.25)2=39.0625 16 16-24.75= -8.75 (-8.75)2=76.5625 21 21-24.75 = -3.75 (-3.75)2=14.0625

The next step is to add the squared values, giving the amount that will be in the numerator of the formula. Adding the values in the third column gives Σ(? − ?̅)2 = 607.5.

Because n=8, the calculation is now: ? = √607.5 8−1 = √607.5 7 = √86.7857… = 9.3158… ≈ 9.3

Note that no rounding is done until the final result, giving a standard deviation of s=9.3.


Part 2 Exercises: For each of the data sets provided below, calculate the mean, median, mode, and standard deviation. Round the answers to one decimal place and show all work.

1. Data Set 1: 26, 31, 10, 37, 38 a. mean = ? b. median = ? c. mode = ? d. standard deviation = ?

2. Data Set 2: 48, 42, 23, 26, 50, 39, 55, 62, 50, 42 a. mean = ? b. median = ? c. mode = ? d. standard deviation = ?


Part 3: Creating Frequency Distributions and Histograms

Example: It is often quite useful to represent data graphically. One popular way to represent data is the histogram. The following set of 16 values can be organized into a frequency distribution (table) and then the frequency distribution is used to create a histogram.

18, 16, 14, 12, 13, 19, 17, 14, 10, 9, 11, 14, 13, 19, 10, 16

The sorted data set is: 9, 10, 10, 11, 12, 13, 13, 14, 14, 14, 16, 16, 17, 18, 19, 19

To create a histogram from this set of data we start by constructing a frequency distribution or table. A frequency distribution shows how data is split up into categories or classes by listing
6

the classes along with the number of data values in each class. The first step in creating a frequency distribution is deciding how many classes we wish to use. For this data set let’s use four classes. Once that decision has been made there is a step-by-step process we can follow.

Step 1: To calculate the class width, find the following value.

????ℎ =
??????? ?????−??????? ????? ?????? ?? ???????


If this value is a decimal, round up to the nearest integer (illustrated in this example). If this value is an integer, you may have to add one in order to have enough classes to accommodate the data.

For the data set listed above, the width is:

width =
19−9 4
= 2.5

This value is rounded up to give a width of 3.

Step 2: Choose a value for the first lower class limit. Typically, this is the minimum value but it could also be a conveniently chosen value.

For the data set listed above, the first lower class limit will be 9.

Step 3: Use the first lower class limit and class width to list the other lower class limits. Add the class width to each lower class limit to determine the next lower class limit.

First lower class limit = 9 Second lower class limit = 9+3 = 12 Third lower class limit = 12+3 = 15 Fourth lower class limit = 15+3 = 18

This creates the start of a frequency distribution table as shown below:







Step 4: Determine the upper class limits. The upper class limit in the FIRST class will be the value just below the lower class limit in the SECOND class. Then add the width to get the remaining upper class limits.

First upper class limit = 11, because 11 is the value right before 12 Second upper class limit = 11+3 = 14 Third upper class limit = 14+3 = 17 Fourth upper class limit = 17+3 = 20
Class Frequency 9 - 12 - 15 - 18 -
7


Now the classes in the table are complete, as shown below.   






Step 5: Use the sorted list to determine how many values belong in each class and enter the value into the Frequency column. Be sure to check that all values are included by adding the frequencies to get the sample size.

9, 10, 10, 11, 12, 13, 13, 14, 14, 14, 16, 16, 17, 18, 19, 19

The first class has a frequency of 4, because 9, 10, 10, 11 are in the first class. The second class has a frequency of 6, because 12, 13, 13, 14, 14, 14 are in the second class. The third class has a frequency of 3, because 16, 16, 17 are in the third class. The fourth class has a frequency of 3, because 18, 19, 19 are in the fourth class.   






For the frequency distribution above, the sum of the frequencies is equal to the sample size: 4 + 6 + 3 + 3 = 16

Step 6: Follow the directions from the link on how to “Create a Histogram From a Frequency Table” (included below) to create a histogram similar to the one shown here.

  

0
1
2
3
4
5
6
7
9 - 11 12 - 14 15 - 17 18 - 20
Frequency
Data Values
Histogram
Class Frequency 9 - 11 12 - 14 15 - 17 18 - 20
Class Frequency 9 - 11 4 12 - 14 6 15 - 17 3 18 - 20 3
8

The following links are for videos demonstrating how to use Excel.

Using Excel for Data and Tables Create a Histogram from a Frequency Table


Part 3 Exercises:

Use the following data set to complete the problems.

15, 26, 31, 10, 37, 38, 35, 30, 24, 21, 26, 27, 24, 32, 32, 19, 34, 32, 24, 39, 30

1. List the data values in order from lowest to highest. 2. Find n, the number of data values. 3. Find the width for a frequency distribution with 6 classes of equal width. 4. Create a frequency distribution table. 5. Show that the frequencies in the table add up to the sample size. 6. Create a histogram.

2. Imagine a collection of nine rooms, R1, R2,...,R9 connected as in the figure below. A social gathering involving N people takes place in these rooms, and the percentage of persons in each room is x1, x2,...,x9. As time progresses it is observed that in a time span of fifteen minutes people mingle and either stay where they are or migrate to an adjacent room with equal probability.

(a) Suppose that 100 people are distributed through out the rooms so that R1 has 30 people, R2 has 40 people, R5 has 10 people and R8 has 20 people. After fifteen minutes how many people do we expect to have in each room.

(b) The number of people observed at a given time in rooms R1, R2,...,R9 respectively is (125,50,150,75,275,100,125,50,150). What was the number of persons in each room fifteen minutes earlier?

(c) If everybody begins in R1, how long until there are people in R9? At that time, what is the distribution of people throughout the rooms?

(d) If you were to cater an event at this location, how would you divide your resources through out the nine rooms?

Recall that Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Now suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 219 numerical entries from the file and r = 48 of the entries had a first nonzero digit of 1.

Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. (i) Test the claim that p is less than 0.301. Use α = 0.10. (a) What is the level of significance?

State the null and alternate hypotheses. a) H0: p = 0.301; H1: p ≠ 0.301 b) H0: p < 0.301; H1: p = 0.301 c) H0: p = 0.301; H1: p > 0.301 d) H0: p = 0.301; H1: p < 0.301 I

Suppose that the average trip time from your house to school on the train is 24 minutes and that the standard deviation of the population is 4 minutes (sigma=4). We are interested to know whether the average trip time on Thursdays is shorter. We study a sample of 49 trips on Thursdays and the average is 22.75 minutes. Execute a hypothesis test at alpha=10% that the average trip time on Thursdays is shorter (that is, that it requires a smaller number of minutes). State Ho, Ha, calculate the appropriate statistic, p-value, state whether you reject Ho vs. not and state your conclusion in plain English.

A) What is the largest number of minutes that will allow you to reject Ho?

B) What is the probability that you will reject Ho if mu is actually 23.375 minutes?

C) Fill in the probabilities in the table below. Also indicate alpha, beta and power. D)What test concludes Ho is true in population Ho not true in population Reject Ho Not reject Ho

D) Fill in the probabilities in the table below. Also indicate alpha, beta and power.

{What test concludes; Ho is true in population ; Ho not true in population }

{Reject Ho ; ; }

{Not reject Ho ; ; }