HackerRank Probability – Normal Distribution #3

Challenge: Normal Distribution #3

Subdomeniu: Probability (probability)

Scor cont: 5.0 / 5

Submission status: Accepted

Submission score: 1.0

Submission ID: 464719021

Limbaj: python3

Link challenge: https://www.hackerrank.com/challenges/normal-distribution-3/problem

Cerinta

A large group of students took a test in physics and the final grades have a mean of 70 and a standard deviation of 10. If we can approximate the distribution of these grades by a normal distribution, what percent of the students 

a) scored higher than 80? 

b) should pass the test (grades≥60)? 

c) should fail the test (grades<60)?

**Submission Modes and Output Format**

You may submit either an R or Python program to accomplish the above task or solve the problem on pen-and-paper. Your output should be three floating point/decimal numbers separated by a line, correct to 2 places of decimal.

1. In the text box below, enter three floating point/decimal numbers, correct to 2 places of decimal.

2. Alternatively, you may submit an R program, which uses the above parameters (hard-coded) and computes the answer.

Your answer should resemble something like:
<pre>
65.12
15.45
2.78
</pre>
(This is **NOT** the answer, just a demonstration of what the answering format should resemble).

Cod sursa

# Enter your code here. Read input from STDIN. Print output to STDOUT
import math
def get_function(x):
    mu = 70
    sig = 10
    gaussian_func = (1 / (sig * math.sqrt(2 * math.pi))) * math.exp(-((x - mu) ** 2) / (2 * sig ** 2))
    return gaussian_func

start_value = 20
end_value = 120
step = 0.0001

x = [start_value + i * step for i in range(int((end_value - start_value) / step) + 1)]
y0 = 0
y1 = 0
y2 = 0
y3 = 0

for i in range(len(x)):
    y0 += get_function(x[i]) * step
    if x[i] > 80:
        y1 += get_function(x[i]) * step

    if x[i] >= 60:
        y2 += get_function(x[i]) * step

    if x[i] < 60:
        y3 += get_function(x[i]) * step

y1 = y1 * 100 / y0
y2 = y2 * 100 / y0
y3 = y3 * 100 / y0

result1 = round(y1, 2)
result2 = round(y2, 2)
result3 = round(y3, 2)

print(str(result1))
print(str(result2))
print(str(result3))