Program for class interval arithmetic mean

Given a class interval and frequency distribution and the task is to find Arithmetic mean. In case of frequency distribution the raw data is arranged by intervals having corresponding frequencies. So if we are interested to find arithmetic mean of the data having class interval we must know the mid variable x. This variable can be calculated by using mid point of interval.

Let lower limit of interval are lower_limit[] = {1, 6, 11, 16, 21} 
Upper limit of interval are upper_limit[] = {5, 10, 15, 20, 25} 
and frequency freq[] = {10, 20, 30, 40, 50} are given.
Where mid(x) = (lower[i] + upper[i]) / 2; 
and mean = (freq[0] * mid[0] + freq[1] * mid[1] + . . . + freq[n – 1] * mid[n – 1]) / (freq[0] + freq[1] + . . . + freq[n-1])
= 2450 / 150 
= 16.3333 

Examples:  

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>

Output

16.3333

Approach#2: Using list comprehension and zip

This approach calculates the class interval arithmetic mean given the lower and upper limits of each class interval and their corresponding frequencies.

Algorithm

1. Calculate the number of observations by summing the frequencies: n = sum(freq)
2. Calculate the midpoint of each class interval by taking the average of the lower and upper limits: xi_list = [(a + b) / 2 for a, b in zip(lower_limit, upper_limit)]
3. Calculate the product of each midpoint and its corresponding frequency: fi_xi_list = [a * b for a, b in zip(freq, xi_list)]
4. Calculate the sum of all products: sum_fi_xi = sum(fi_xi_list)
5. Calculate the arithmetic mean by dividing the sum of all products by the total number of observations: arithmetic_mean = sum_fi_xi / n
6. Output the result: print(“Class interval arithmetic mean is:”, arithmetic_mean)

Python3

lower_limit = [10, 20, 30, 40, 50]
upper_limit = [19, 29, 39, 49, 59]
freq = [15, 20, 30, 35, 40]

n = sum(freq)
xi_list = [(a + b) / 2 for a, b in zip(lower_limit, upper_limit)]
fi_xi_list = [a * b for a, b in zip(freq, xi_list)]
arithmetic_mean = sum(fi_xi_list) / n

print("Class interval arithmetic mean is:", arithmetic_mean)

Output

Class interval arithmetic mean is: 39.142857142857146

Time complexity: O(n), where n is the number of class intervals. This is because the code performs a single loop over the class intervals to calculate the midpoints, and a second loop to calculate the products.

Space complexity: O(n), as it uses three lists to store the lower limits, upper limits, and frequencies of each class interval, as well as two additional lists for the midpoints and products. However, these additional lists are relatively small compared to the original input lists, so the overall space complexity is still O(n).