Which sentence describes a task on which a genetic engineer is most likely to work?
2 answers:
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Answer:
Check the explanation
Explanation:
<u>The Code</u>
#include <fstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
//Function Declarations
void fillArrayWithRandNos(int nos[],int size);
void displayArray(int nos[],int size);
double mean(int nos[],int size);
double variance(int nos[],int size);
double median(int nos[],int size);
int mode(int nos[],int size);
void histogram(int nos[],int size);
int main() {
//Declaring variables
const int size=100;
srand(time(NULL));
// Creating array dynamically
int* nos = new int[size];
//Calling the functions
fillArrayWithRandNos(nos,size);
displayArray(nos,size);
cout<<"Mean :"<<mean(nos,size)<<endl;
cout<<"variance :"<<variance(nos,size)<<endl;
cout<<"Median :"<<median(nos,size)<<endl;
cout<<"Mode :"<<mode(nos,size)<<endl;
histogram(nos,size);
return 0;
}
void fillArrayWithRandNos(int nos[],int size)
{
for(int i=0;i<size;i++)
{
nos[i]=rand()%(45) + 55;
}
}
void displayArray(int nos[],int size)
{
cout<<"Displaying the array elements :"<<endl;
for(int i=0;i<size;i++)
{
cout<<nos[i]<<" ";
if((i+1)%10==0)
cout<<endl;
}
}
double mean(int nos[],int size)
{
double sum=0;
for(int i=0;i<size;i++)
{
sum+=nos[i];
}
return sum/size;
}
double variance(int nos[],int size)
{
double avg=mean(nos,size);
double variance,sum=0;
for(int i=0;i<size;i++)
{
sum+=pow(nos[i]-avg,2);
}
//calculating the standard deviation of nos[] array
variance=(double)sum/(size);
return variance;
}
double median(int nos[],int size)
{
//This Logic will Sort the Array of elements in Ascending order
int temp;
for (int i = 0; i < size; i++)
{
for (int j = i + 1; j < size; j++)
{
if (nos[i] > nos[j])
{
temp = nos[i];
nos[i] = nos[j];
nos[j] = temp;
}
}
}
int middle;
float med;
middle = (size / 2.0);
if (size % 2 == 0)
med = ((nos[middle - 1]) + (nos[middle])) / 2.0;
else
med = (nos[middle]);
return med;
}
int mode(int nos[],int size)
{
int counter1 = 0, counter2, modevalue;
for (int i = 0; i < size; i++) {
counter2 = 0;
for (int j = i; j < size; j++) {
if (nos[i] == nos[j]) {
counter2++;
}
if (counter2 > counter1) {
counter1 = counter2;
modevalue = nos[i];
}
}
}
if (counter1 > 1)
return modevalue;
else
return 0;
}
void histogram(int nos[],int size)
{
int hist[9]={0};
for(int i=0;i<size;i++)
{
if(nos[i]>=55 && nos[i]<=59)
{
hist[0]++;
}
else if(nos[i]>=60 && nos[i]<=64)
{
hist[1]++;
}
else if(nos[i]>=65 && nos[i]<=69)
{
hist[2]++;
}
else if(nos[i]>=70 && nos[i]<=74)
{
hist[3]++;
}
else if(nos[i]>=75 && nos[i]<=79)
{
hist[4]++;
}
else if(nos[i]>=80 && nos[i]<=84)
{
hist[5]++;
}
else if(nos[i]>=85 && nos[i]<=89)
{
hist[6]++;
}
else if(nos[i]>=90 && nos[i]<=94)
{
hist[7]++;
}
else if(nos[i]>=95 && nos[i]<=99)
{
hist[8]++;
}
}
cout<<"Displaying the count of numbers in each interval:"<<endl;
int cnt=0;
for(int i=55;i<=99;i+=5)
{
cout<<i<<"-"<<i+4<<"|"<<hist[cnt]<<endl;
cnt++;
}
cout<<"Displaying the histogram :"<<endl;
cnt=0;
for(int i=55;i<=99;i+=5)
{
cout<<i<<"-"<<i+4<<"|";
for(int j=0;j<hist[cnt];j++)
{
cout<<"*";
}
cout<<endl;
cnt++;
}
}
#########
___________________________
The output can be seen in the attached image below.
