Hello coders, today we are going to solve **Day 5: Normal Distribution II HackerRank Solution** which is a Part of **10 Days of Statistics** Series.

**Objective**

In this challenge, we go further with normal distributions.

**Task**

The final grades for a Physics exam taken by a large group of students have a mean of ** U = 70** and a standard deviation of

**. If we can approximate the distribution of these grades by a normal distribution, what percentage of the students:**

*O*= 10- Scored higher than
**80**(i.e., have a)?*grade*> 80 - Passed the test (i.e., have a
)?*grade*=> 60 - Failed the test (i.e., have a
)?*grade*< 60

Find and print the answer to each question on a new line, rounded to a scale of **2** decimal places.

**Input Format**

There are **3** lines of input (shown below):

```
70 10
80
60
```

The first line contains **2** space-separated values denoting the respective mean and standard deviation for the exam. The second line contains the number associated with question **1**. The third line contains the pass/fail threshold number associated with questions **2** and **3**.

If you do not wish to read this information from stdin, you can hard-code it into your program.

**Output Format**

There are three lines of output. Your answers must be rounded to a scale of **2** decimal places (i.e., **1.23** format):

- On the first line, print the answer to question
**1**(i.e., the percentage of students having).*grade*> 80 - On the second line, print the answer to question
**2**(i.e., the percentage of students having).*grade*=> 60 - On the third line, print the answer to question
**3**(i.e., the percentage of students having).*grade*< 60

**Solution – Normal Distribution II**

**C++**

#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; double normal_dist(double m, double sd, double x) { double p = 0.5*(1 + erf((x-m)/(sd*sqrt(2.0)))); return p*100; // for percentage } int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ double m = 70, sd = 10; double p1 = double(100) - normal_dist(m, sd, 80); double p2 = double(100) - normal_dist(m, sd, 60); double p3 = normal_dist(m, sd, 60); printf("%0.2f\n%0.2f\n%0.2f", p1, p2, p3); return 0; }

**Disclaimer: **The above Problem **(Normal Distribution II)** is generated by **Hacker Rank **but the Solution is Provided by **CodingBroz**. This tutorial is only for** Educational** and **Learning** Purpose.