06.03.2022

what is the bin size of this histogram?

. 4

Faq

Mathematics
Step-by-step answer
P Answered by Specialist

The frequency distribution and percent frequency distribution are given in the explanation below. The histogram is attached as an image.

Step-by-step explanation:

U.S. Box Office Receipts (Inflation Adjusted Millions $)

$1,650 $1,426 $1,145 $1,132 $1,096 $1,053 $1,029 $973 $871 $854 $844 $825 $809 $804 $772 $741 $722 $720 $715 $686 $683 $676 $671 $651 $623 $623 $618 $612 $589 $570 $562 $557 $553 $552 $552 $549 $549 $548 $529 $528 $527 $518 $515 $515 $513 $513 $507 $496 $496 $494

Using bin sizes of $100 Million, we can create the frequency distribution:

Inflation Adjusted Million $             Frequency

$400≤x<$500                                          3

$500≤x<$600                                         19

$600≤x<$700                                           9

$700≤x<$800                                           5

$800≤x<$900                                           6

$900≤x<$1000                                          1

$1000≤x<$1100                                          3

$1100≤x<$1200                                          2

$1200≤x<$1300                                         0

$1300≤x<$1400                                         0

$1400≤x<$1500                                          1

$1500≤x<$1600                                         0

$1600≤x<$1700                                          1

To create a percent frequency distribution, we will use the following formula for each class:

(frequency/total frequency) * 100

Total frequency = 50.

Inflation Adjusted Million $             Percentage Frequency(%)

$400≤x<$500                                          3/50 * 100 = 6

$500≤x<$600                                         19/50 * 100 = 38

$600≤x<$700                                           9/50 * 100 = 18

$700≤x<$800                                           5/50 * 100 = 10

$800≤x<$900                                           6/50 * 100 = 12

$900≤x<$1000                                          1/50 * 100 = 2

$1000≤x<$1100                                          3/50 * 100 = 6

$1100≤x<$1200                                          2/50 * 100 = 4

$1200≤x<$1300                                         0/50 * 100 = 0

$1300≤x<$1400                                         0/50 * 100 = 0

$1400≤x<$1500                                          1/50 * 100 = 2

$1500≤x<$1600                                         0/50 * 100 = 0

$1600≤x<$1700                                          1/50 * 100 = 2

We can create a histogram using the percent frequency distribution values. Plot the percent frequencies on the y-axis and the inflation adjusted on the x-axis. I am attaching the histogram as an image here.


Create a frequency distribution, percent frequency distribution, and histogram for inflation-adjuste
Mathematics
Step-by-step answer
P Answered by Master

The frequency distribution and percent frequency distribution are given in the explanation below. The histogram is attached as an image.

Step-by-step explanation:

U.S. Box Office Receipts (Inflation Adjusted Millions $)

$1,650 $1,426 $1,145 $1,132 $1,096 $1,053 $1,029 $973 $871 $854 $844 $825 $809 $804 $772 $741 $722 $720 $715 $686 $683 $676 $671 $651 $623 $623 $618 $612 $589 $570 $562 $557 $553 $552 $552 $549 $549 $548 $529 $528 $527 $518 $515 $515 $513 $513 $507 $496 $496 $494

Using bin sizes of $100 Million, we can create the frequency distribution:

Inflation Adjusted Million $             Frequency

$400≤x<$500                                          3

$500≤x<$600                                         19

$600≤x<$700                                           9

$700≤x<$800                                           5

$800≤x<$900                                           6

$900≤x<$1000                                          1

$1000≤x<$1100                                          3

$1100≤x<$1200                                          2

$1200≤x<$1300                                         0

$1300≤x<$1400                                         0

$1400≤x<$1500                                          1

$1500≤x<$1600                                         0

$1600≤x<$1700                                          1

To create a percent frequency distribution, we will use the following formula for each class:

(frequency/total frequency) * 100

Total frequency = 50.

Inflation Adjusted Million $             Percentage Frequency(%)

$400≤x<$500                                          3/50 * 100 = 6

$500≤x<$600                                         19/50 * 100 = 38

$600≤x<$700                                           9/50 * 100 = 18

$700≤x<$800                                           5/50 * 100 = 10

$800≤x<$900                                           6/50 * 100 = 12

$900≤x<$1000                                          1/50 * 100 = 2

$1000≤x<$1100                                          3/50 * 100 = 6

$1100≤x<$1200                                          2/50 * 100 = 4

$1200≤x<$1300                                         0/50 * 100 = 0

$1300≤x<$1400                                         0/50 * 100 = 0

$1400≤x<$1500                                          1/50 * 100 = 2

$1500≤x<$1600                                         0/50 * 100 = 0

$1600≤x<$1700                                          1/50 * 100 = 2

We can create a histogram using the percent frequency distribution values. Plot the percent frequencies on the y-axis and the inflation adjusted on the x-axis. I am attaching the histogram as an image here.


Create a frequency distribution, percent frequency distribution, and histogram for inflation-adjuste
Computers and Technology
Step-by-step answer
P Answered by PhD

import java.util.*;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

class GFG

{

  // Function for calculating mean

  public static double findMean(double a[], int n)

  {

      int sum = 0;

      for (int i = 0; i < n; i++)

          sum += a[i];

 

      return (double)sum / (double)n;

  }

  // Function for calculating median

  public static double findMedian(double a[], int n)

  {

      // First we sort the array

      Arrays.sort(a);

      // check for even case

      if (n % 2 != 0)

      return (double)a[n / 2];

 

      return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;

  }

  public static double findMode(double a[], int n)

{

// The output array b[] will

// have sorted array

//int []b = new int[n];

 

// variable to store max of

// input array which will

// to have size of count array

double max = Arrays.stream(a).max().getAsDouble();

 

// auxiliary(count) array to

// store count. Initialize

// count array as 0. Size

// of count array will be

// equal to (max + 1).

double t = max + 1;

double[] count = new double[(int)t];

for (int i = 0; i < t; i++)

{

count[i] = 0;

}

 

// Store count of each element

// of input array

for (int i = 0; i < n; i++)

{

count[(int)(10*a[i])]++;

}

 

// mode is the index with maximum count

double mode = 0;

double k = count[0];

for (int i = 1; i < t; i++)

{

if (count[i] > k)

{

k = count[i];

mode = i;

}

}

return mode;

}

public static double findSmallest(double [] A, int total){

Arrays.sort(A);

return A[0];

}

 

public static void printAboveAvg(double arr[], int n)

{

 

// Find average

double avg = 0;

for (int i = 0; i < n; i++)

avg += arr[i];

avg = avg / n;

 

// Print elements greater than average

for (int i = 0; i < n; i++)

if (arr[i] > avg)

System.out.print(arr[i] + " ");

System.out.println();

}

 

public static void printrand(double [] A, int n){

Arrays.sort(A);

for(int i=0;i<n;i++){

System.out.print(A[0]+"/t");

}

System.out.println();

}

 

public static void printHist(double [] arr, int n) {

 

for (double i = 1.0; i >= 0; i-=0.1) {

System.out.print(i+" | ");

for (int j = 0; j < n; j++) {

 

// if array of element is greater

// then array it print x

if (arr[j] >= i)

System.out.print("x");

 

// else print blank spaces

else

System.out.print(" ");

}

System.out.println();

}

// print last line denoted by

for(int l = 0; l < (n + 3); l++){    

System.out.print("---");

}

 

System.out.println();

System.out.print(" ");

 

for (int k = 0; k < n; k++) {

System.out.print(arr[k]+" ");

}

}

  // Driver program

  public static void main(String args[]) throws IOException

  {

      //Enter data using BufferReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

double [] A = new double[100];

int i=0;

System.out.println("Enter the numbers(0.0-1.0) /n Enter 9 if u have entered the numbers. /n");

do

{

A[i++]=Double.parseDouble(br.readLine());

}while(A[i-1]==9);

      i--;

      System.out.println("Average = " + findMean(A,i) );

      System.out.println("Median = " + findMedian(A,i));

      System.out.println("Element that occured most frequently = " + findMode(A,i));

      System.out.println("number closest to 0.0 =" + findSmallest(A,i));

      System.out.println("Numbers that are greater than the average are follows:");

      printAboveAvg(A,i);

      System.out.println("Numbers in random order are as follows:");

      printrand(A,i);

      System.out.println("Histogram is bellow:");

      printHist(A,i);

  }

}

Explanation:

Computers and Technology
Step-by-step answer
P Answered by PhD

import java.util.*;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

class GFG

{

  // Function for calculating mean

  public static double findMean(double a[], int n)

  {

      int sum = 0;

      for (int i = 0; i < n; i++)

          sum += a[i];

 

      return (double)sum / (double)n;

  }

  // Function for calculating median

  public static double findMedian(double a[], int n)

  {

      // First we sort the array

      Arrays.sort(a);

      // check for even case

      if (n % 2 != 0)

      return (double)a[n / 2];

 

      return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;

  }

  public static double findMode(double a[], int n)

{

// The output array b[] will

// have sorted array

//int []b = new int[n];

 

// variable to store max of

// input array which will

// to have size of count array

double max = Arrays.stream(a).max().getAsDouble();

 

// auxiliary(count) array to

// store count. Initialize

// count array as 0. Size

// of count array will be

// equal to (max + 1).

double t = max + 1;

double[] count = new double[(int)t];

for (int i = 0; i < t; i++)

{

count[i] = 0;

}

 

// Store count of each element

// of input array

for (int i = 0; i < n; i++)

{

count[(int)(10*a[i])]++;

}

 

// mode is the index with maximum count

double mode = 0;

double k = count[0];

for (int i = 1; i < t; i++)

{

if (count[i] > k)

{

k = count[i];

mode = i;

}

}

return mode;

}

public static double findSmallest(double [] A, int total){

Arrays.sort(A);

return A[0];

}

 

public static void printAboveAvg(double arr[], int n)

{

 

// Find average

double avg = 0;

for (int i = 0; i < n; i++)

avg += arr[i];

avg = avg / n;

 

// Print elements greater than average

for (int i = 0; i < n; i++)

if (arr[i] > avg)

System.out.print(arr[i] + " ");

System.out.println();

}

 

public static void printrand(double [] A, int n){

Arrays.sort(A);

for(int i=0;i<n;i++){

System.out.print(A[0]+"/t");

}

System.out.println();

}

 

public static void printHist(double [] arr, int n) {

 

for (double i = 1.0; i >= 0; i-=0.1) {

System.out.print(i+" | ");

for (int j = 0; j < n; j++) {

 

// if array of element is greater

// then array it print x

if (arr[j] >= i)

System.out.print("x");

 

// else print blank spaces

else

System.out.print(" ");

}

System.out.println();

}

// print last line denoted by

for(int l = 0; l < (n + 3); l++){    

System.out.print("---");

}

 

System.out.println();

System.out.print(" ");

 

for (int k = 0; k < n; k++) {

System.out.print(arr[k]+" ");

}

}

  // Driver program

  public static void main(String args[]) throws IOException

  {

      //Enter data using BufferReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

double [] A = new double[100];

int i=0;

System.out.println("Enter the numbers(0.0-1.0) /n Enter 9 if u have entered the numbers. /n");

do

{

A[i++]=Double.parseDouble(br.readLine());

}while(A[i-1]==9);

      i--;

      System.out.println("Average = " + findMean(A,i) );

      System.out.println("Median = " + findMedian(A,i));

      System.out.println("Element that occured most frequently = " + findMode(A,i));

      System.out.println("number closest to 0.0 =" + findSmallest(A,i));

      System.out.println("Numbers that are greater than the average are follows:");

      printAboveAvg(A,i);

      System.out.println("Numbers in random order are as follows:");

      printrand(A,i);

      System.out.println("Histogram is bellow:");

      printHist(A,i);

  }

}

Explanation:

Mathematics
Step-by-step answer
P Answered by Master

A) 0.028

Step-by-step explanation:

Given:

Sample size, n = 115

Population parameter, p = 0.1

The X-Bin(n=155, p=0.1)

Required:

Find the standard deviation of the normal curve that can be used to approximate the binomial probability histogram.

To find the standard deviation, use the formula below:

\sigma = \sqrt{\frac{p(1-p)}{n}}

Substitute figures in the equation:

\sigma = \sqrt{\frac{0.1(1 - 0.1)}{115}}

\sigma = \sqrt{\frac{0.1 * 0.9}{115}}

\sigma = \sqrt{\frac{0.09}{115}}

\sigma = \sqrt{7.826*10^-^4}

\sigma = 0.028

The Standard deviation of the normal curve that can be used to approximate the binomial probability histogram is 0.028

Try asking the Studen AI a question.

It will provide an instant answer!

FREE