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  1. Prepare
  2. Mathematics
  3. Probability
  4. Normal Distribution #3
  5. Discussions

Normal Distribution #3

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15 Discussions

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  • gschoen
     + 0 comments

    Why is this a hard question? You can get the answers straight out of Excel.

    Incidentally, each student's score should be an integer and not a real number (as assumed in this question).

  • addsc7
     + 0 comments

    Answer:

    # 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))

  • cjckris
     + 0 comments

    from math import exp, factorial, pi, erf

    Cumulative Probability

    Probility that in a normal distribution the value is less than x.

    def CDF(x, mu, std): return 0.5*(1+erf((x-mu)/(std*((2**0.5)))))

  • Mandar_Inamdar
     + 0 comments

    Python code I used successfully:

    from scipy.stats import norm
a=100*(1-norm.cdf(80, loc=70, scale=10))
b=100*norm.sf(60, loc=70, scale=10)
c=100*norm.cdf(60, loc=70, scale=10)

print "%.2f" %a
print "%.2f" %b
print "%.2f" %c

  • ksvtu5379
     + 0 comments

    is thr ny negative value for (b)part??