Dado un intervalo de clase y una distribución de frecuencias, la tarea es encontrar la media aritmética. En el caso de la distribución de frecuencias, los datos sin procesar se organizan por intervalos que tienen frecuencias correspondientes. Entonces, si estamos interesados en encontrar la media aritmética de los datos que tienen un intervalo de clase, debemos conocer la variable media x. Esta variable se puede calcular utilizando el punto medio del intervalo.
Deje que el límite inferior del intervalo sea límite_inferior[] = {1, 6, 11, 16, 21}
El límite superior del intervalo sea límite_superior[] = {5, 10, 15, 20, 25}
y la frecuencia freq[] = {10, 20, 30, 40, 50} se dan.
Donde mid(x) = (inferior[i] + superior[i]) / 2;
y media = (freq[0] * mid[0] + freq[1] * mid[1] + . . . + freq[n – 1] * mid[n – 1]) / (freq[0] + freq [1] + . . . + frecuencia[n-1])
= 2450 / 150
= 16,3333
Ejemplos:
Input : lower_limit[] = {1, 6, 11, 16, 21}
upper_limit[] = {5, 10, 15, 20, 25}
freq[] = {10, 20, 30, 40, 50}
Output : 16.3333
Input : lower_limit[] = {10, 20, 30, 40, 50}
upper_limit[] = {19, 29, 39, 49, 59}
freq[] = {15, 20, 30, 35, 40}
Output : 38.6429
C++
// CPP program to find class interval
// arithmetic mean.
#include <bits/stdc++.h>
using namespace std;
// Function to find class interval arithmetic mean.
float mean(int lower_limit[], int upper_limit[],
int freq[], int n)
{
float mid[n];
float sum = 0, freqSum = 0;
// calculate sum of frequency and sum of
// multiplication of interval mid value
// and frequency.
for (int i = 0; i < n; i++) {
mid[i] = (lower_limit[i] +
upper_limit[i]) / 2;
sum = sum + mid[i] * freq[i];
freqSum = freqSum + freq[i];
}
return sum / freqSum;
}
// Driver function
int main()
{
int lower_limit[] = { 1, 6, 11, 16, 21 };
int upper_limit[] = { 5, 10, 15, 20, 25 };
int freq[] = { 10, 20, 30, 40, 50 };
int n = sizeof(freq) / sizeof(freq[0]);
cout << mean(lower_limit, upper_limit, freq, n);
return 0;
}
Java
// java program to find
// class interval
import java.io.*;
class GFG {
// Function to find class
// interval arithmetic mean.
static float mean(int lower_limit[],
int upper_limit[], int freq[], int n)
{
float mid[] = new float[n];
float sum = 0, freqSum = 0;
// calculate sum of frequency and sum of
// multiplication of interval mid value
// and frequency.
for (int i = 0; i < n; i++) {
mid[i] = (lower_limit[i] +
upper_limit[i]) / 2;
sum = sum + mid[i] * freq[i];
freqSum = freqSum + freq[i];
}
return sum / freqSum;
}
// Driver function
public static void main (String[] args) {
int lower_limit[] = { 1, 6, 11, 16, 21 };
int upper_limit[] = { 5, 10, 15, 20, 25 };
int freq[] = { 10, 20, 30, 40, 50 };
int n = freq.length;
mean(lower_limit, upper_limit, freq, n);
System.out.println(mean(lower_limit,
upper_limit, freq, n));
}
}
// This code is contributed by vt_m
Python3
# Python 3 program to find class interval # arithmetic mean. # Function to find class interval # arithmetic mean. def mean(lower_limit, upper_limit, freq, n): mid = [0.0] * n sum = 0 freqSum = 0 # calculate sum of frequency and # sum of multiplication of interval # mid value and frequency. for i in range( 0, n): mid[i] = ((lower_limit[i] + upper_limit[i]) / 2) sum = sum + mid[i] * freq[i] freqSum = freqSum + freq[i] return sum / freqSum # Driver function lower_limit = [ 1, 6, 11, 16, 21 ] upper_limit = [ 5, 10, 15, 20, 25 ] freq = [10, 20, 30, 40, 50] n = len(freq) print(round(mean(lower_limit, upper_limit, freq, n), 4)) # This code is contributed by # Smitha Dinesh Semwal
C#
// C# program to find
// class interval
using System;
class GFG {
// Function to find class
// interval arithmetic mean.
static float mean(int []lower_limit,
int []upper_limit, int []freq, int n)
{
float []mid = new float[n];
float sum = 0, freqSum = 0;
// calculate sum of frequency and sum of
// multiplication of interval mid value
// and frequency.
for (int i = 0; i < n; i++) {
mid[i] = (lower_limit[i] +
upper_limit[i]) / 2;
sum = sum + mid[i] * freq[i];
freqSum = freqSum + freq[i];
}
return sum / freqSum;
}
// Driver function
public static void Main () {
int []lower_limit = { 1, 6, 11, 16, 21 };
int []upper_limit = { 5, 10, 15, 20, 25 };
int []freq = { 10, 20, 30, 40, 50 };
int n = freq.Length;
mean(lower_limit, upper_limit, freq, n);
Console.WriteLine(mean(lower_limit,
upper_limit, freq, n));
}
}
// This code is contributed by vt_m
PHP
<?php
// PHP program to find class interval
// arithmetic mean.
// Function to find class interval
// arithmetic mean.
function mean( $lower_limit, $upper_limit,
$freq, $n)
{
$mid = array();
$sum = 0; $freqSum = 0;
// calculate sum of frequency and
// sum of multiplication of interval
// mid value and frequency.
for ( $i = 0; $i <$n; $i++)
{
$mid[$i] = ($lower_limit[$i] +
$upper_limit[$i]) / 2;
$sum = $sum + $mid[$i] * $freq[$i];
$freqSum = $freqSum + $freq[$i];
}
return $sum / $freqSum;
}
// Driver function
$lower_limit = array( 1, 6, 11, 16, 21 );
$upper_limit = array( 5, 10, 15, 20, 25 );
$freq = array( 10, 20, 30, 40, 50 );
$n = count($freq);
echo mean($lower_limit, $upper_limit,
$freq, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program to find
// class interval
// Function to find class
// interval arithmetic mean.
function mean(lower_limit,
upper_limit, freq, n)
{
let mid = [];
let sum = 0, freqSum = 0;
// calculate sum of frequency and sum of
// multiplication of interval mid value
// and frequency.
for (let i = 0; i < n; i++) {
mid[i] = (lower_limit[i] +
upper_limit[i]) / 2;
sum = sum + mid[i] * freq[i];
freqSum = freqSum + freq[i];
}
return sum / freqSum;
}
// Driver Code
let lower_limit = [ 1, 6, 11, 16, 21 ];
let upper_limit = [ 5, 10, 15, 20, 25 ];
let freq = [ 10, 20, 30, 40, 50 ];
let n = freq.length;
mean(lower_limit, upper_limit, freq, n);
document.write(mean(lower_limit,
upper_limit, freq, n));
// This code is contributed by chinmoy1997pal.
</script>
Producción:
16.3333
Publicación traducida automáticamente
Artículo escrito por Dharmendra_Kumar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA