For this challenge I used python programming. To challenge myself I have not imported any python packages other than essential math functions from math import exp, sqrt, pi, erf. There are some multiple choice questions in the challenge, But in this website I have only included the solutions of coding questions. To make learning seamless I have added the statistical equation used to solve the problem. Happy learning!!
Table of contects
Day 0: Mean, Median, and Mode
\[mean= \mu = { \sum_{i=1}^n x_{i}\over n}\] \[\quad n = \textrm{number of values in data set}\] \[median = \left\{ \begin{array}{ c l } X[\frac{n}{2}] & \quad \textrm{if n is even}\\ {(X[\frac{n-1}{2}] + X[\frac{n+1}{2}])\over 2}& \quad \textrm{if n is odd} \end{array} \right.\] \[X = \textrm{ordered list of values in data set}\] \[mode = \textrm{a number that appears most frequently in a data set}\]
N = int(input())
sample = sorted(list(map(float, input().strip().split())))
if N == len(sample):
# mean
sum_ = 0
for i in sample:
sum_ += i
mean = sum_/N
# median
if N%2 == 0:
median = (sample[int(N/2-0.5)] + sample[int(N/2+0.5)])/2
else:
median = sample[int(N/2+0.5)]
# mode
mode = []
count = {}
for i in sample:
if i in count:
count[i] += 1
else:
count[i] = 1
for key in count.keys():
if count[key] == max(count.values()):
mode.append(key)
print(round(mean, 1))
print(round(median, 1))
print(mode[0])
Day 0: Weighted Mean
\[\textrm{Weighted Mean} = m_w = {\sum_{i=1}^n (x_i \times w_i)\over \sum_{i=1}^n w_i}\]def weightedMean(X, W):
global n
XW_sum = 0
W_sum = 0
for i in range(n):
XW_sum += X[i] * W[i]
W_sum += W[i]
print(round((XW_sum/W_sum), 1))
if __name__ == '__main__':
n = int(input().strip())
vals = list(map(int, input().rstrip().split()))
weights = list(map(int, input().rstrip().split()))
weightedMean(vals, weights)
Day 1: Quartiles
\[\textrm{Lower Quartile} =Q_1 = median(X_{1}^{n/2})\] \[\textrm{Middle Quartile} =Q_2 = median(X_{1}^{n})\] \[\textrm{Upper Quartile} =Q_3 = median(X_{n/2}^{n})\] \[X = \textrm{ordered list of values in data set}\\ X_i^j = \textrm{range of X values in between i and j}\\ whereas\; i < j,\; 1 \leq i \leq n,\; 1 \leq j \leq n\]import os
def medium(arr):
n = len(arr)
if n%2 == 0:
Q = (arr[int((n-1)*0.5)] + arr[int((n+1)*0.5)])/2
else:
Q = arr[int(n*0.5)]
if Q - int(Q) == 0:
return int(Q)
else:
return Q
def quartiles(arr):
n = len(arr)
Q2 = medium(arr)
Q1 = medium(arr[:n//2])
Q3 = medium(arr[int(n/2+0.5):])
return Q1, Q2, Q3
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
n = int(input().strip())
data = list(sorted(map(int, input().rstrip().split())))
res = quartiles(data)
fptr.write('\n'.join(map(str, res)))
fptr.write('\n')
fptr.close()
Day 1: Standard Deviation
\[\textrm{Standard Deviation} = \sigma = \sqrt{\sum_{i=1}^n (x_i - \mu)^2\over n}\]def mean(arr):
global n
sum_ = 0
for i in range(n):
sum_ += arr[i]
return n, sum_/n
def stdDev(arr):
n, mean_ = mean(arr)
std = 0
for i in range(n):
std += ((mean_ - arr[i])**2)
print((std/n)**0.5)
if __name__ == '__main__':
n = int(input().strip())
vals = list(map(int, input().rstrip().split()))
stdDev(vals)
Day 1: Interquartile Range
\[\textrm{Interquartile Range} = Q3 - Q1\]def median(arr):
N = len(arr)
if N%2==0:
med = (arr[int((N-1)*0.5)] + arr[int((N+1)*0.5)])/2
else:
med = arr[int(N*0.5)]
return float(med)
def interQuartile(values, freqs):
global n
value_list = []
for i in range(n):
for _ in range(freqs[i]):
value_list.append(values[i])
value_list.sort()
n = len(value_list)
print(round(\
median(value_list[int((n+1)*0.5):]) - median(value_list[:n//2]), 1))
if __name__ == '__main__':
n = int(input().strip())
val = list(map(int, input().rstrip().split()))
freq = list(map(int, input().rstrip().split()))
interQuartile(val, freq)
Day 4: Binomial Distribution I
\[b(x,n,p) ={n! \over x!(n-x)!} \cdot p^x \cdot q^{(n-x)}\] \[b(x \geq r, n, p) = \sum_{i=r}^n b(x=i,n,p)\]p1, n = map(float, input().split())
# p for boys
gap = 1/(1 + (n/p1))
x = 3
n = 6
def fact(x):
a = 1
for i in range(1, x+1):
a *= i
return a
def comb(n, r):
return float(fact(n)/(fact(r)*fact(n-r)))
def bino(x, n, p):
q = 1 - p
return (comb(n, x)*(p**x)*(q**(n-x)))
print(round(sum([bino(i, n, gap) for i in range(x, n+1)]), 3))
Day 4: Binomial Distribution II
\[P(i < x \leq j)= b(i \leq x \leq j, n, p) = \sum_{i=r}^j b(x=i,n,p)\]p, n = map(int, input().split())
p /= 100
def fact(x):
if x == 0:
a = 1
else:
a = 1
for i in range(1, x+1):
a *= i
return a
def comb(n, r):
return float(fact(n)/(fact(r)*fact(n-r)))
def bino(x, n, p):
q = 1 - p
return (comb(n, x)*(p**x)*(q**(n-x)))
def prob(from_, to_):
print(round(sum([bino(i, n, p) for i in range(from_, to_+1)]), 3))
# no more than 2 rejects
prob(0, 2)
# at least 2 rejects
prob(2, n)
Day 4: Geometric Distribution I
\[g(n,p) = q^{n-1} \cdot p\] \[whereas,\;q = 1-p\]p1, p2 = map(int, input().split())
p0 = int(input())
p = p1/p2
def geo(n, p):
return round((p*((1-p)**(n-1))), 3)
print(geo(p0, p))
Day 4: Geometric Distribution II
\[P(x \leq j)= g(n \leq j, p) = \sum_{i=1}^j g(n=i,p)\]p1, p2 = map(int, input().split())
p0 = int(input())
p = p1/p2
def geo(n, p):
return round(sum([(p*((1-p)**(i-1))) for i in range(1, n+1)]), 3)
print(geo(p0, p))
Day 5: Poisson Distribution I
\[P(k, \lambda) = {\lambda^k e^{-\lambda}\over k!}\]from math import exp
mean = float(input())
X = int(input())
def fact(x):
if x == 0:
n = 1
else:
n = 1
for i in range(1, x+1):
n *= i
return n
def pois(X, mean):
return round(((mean**X)*(exp(-mean)))/fact(X), 3)
print(pois(X, mean))
Day 5: Poisson Distribution II
\[E[X^2] = \lambda + \lambda^2\]X, Y = map(float, input().split())
#def C_A(X):
print(round(160 + (40*(X+X**2)), 3))
#def C_B(Y):
print(round(128 + (40*(Y+Y**2)), 3))
Day 5: Normal Distribution I
\[P(X \leq x) = F_X(x) = {1\over2}\Big(1 + erf\Big({x-\mu\over{\sigma \sqrt2}}\Big)\Big)\] \[P(a \leq X \leq b) = F_X(b) - F_X(a)\]from math import exp, sqrt, pi, erf
# read input
mean, std = map(float, input().split())
X1 = float(input())
X2, X3 = map(float, input().split())
'''
# pdf: normal distribution
def norm(x, mean, std):
con1 = exp(-((x - mean)**2)/(2*(std**2)))
return con1/(std*sqrt(2*pi))
'''
# cdf: normal distribution
def norm(x, mean, std):
return ((1 + erf((x - mean)/(sqrt(2)*std)))*0.5)
# Q01
print(round(norm(X1, mean, std), 3))
# Q02
print(round((norm(X3, mean, std) - norm(X2, mean, std)), 3))
Day 5: Normal Distribution II
\[P(X \leq x) = F_X(x)\] \[P(X > x) = 1 - P(X \leq x)= 1 - F_X(x)\]from math import sqrt, erf
mu, std = map(float, input().split())
X1 = float(input())
X2 = float(input())
# cdf: normal distribution
def norm(x, mean, std):
return ((1 + erf((x - mean)/(sqrt(2)*std)))*0.5)*100
# X > X1
print(round(100 - norm(X1, mu, std), 2))
# X >= X2
print(round(100 - norm(X2, mu, std), 2))
# X < X2
print(round(norm(X2, mu, std), 2))
Day 6: The Central Limit Theorem I
\[\mu^\prime = n \times \mu\] \[\sigma^\prime = \sqrt n \times \sigma\] \[F_X(x, \mu^\prime, \sigma^\prime) = {1\over2}\Big(1 + erf\Big({x-\mu^\prime\over{\sigma^\prime \sqrt2}}\Big)\Big)\]from math import sqrt, erf
# read inputs
max_ = int(input())
box = int(input())
mu = int(input())
std = int(input())
# cdf: normal distribution
def norm(x, mean, std):
return ((1 + erf((x - mean)/(sqrt(2)*std)))*0.5)
mu_ = box*mu
std_ = sqrt(box)*std
print(round(norm(max_, mu_, std_), 4))
Day 6: The Central Limit Theorem II
\[\mu^\prime = n \times \mu\] \[\sigma^\prime = \sqrt n \times \sigma\] \[F_X(x, \mu^\prime, \sigma^\prime) = {1\over2}\Big(1 + erf\Big({x-\mu^\prime\over{\sigma^\prime \sqrt2}}\Big)\Big)\]from math import sqrt, erf
# read all inputs
max_tic = int(input())
tic = int(input())
mu = float(input())
std = float(input())
# CDF: normal distribution
def norm(x, mean, std):
return ((1 + erf((x - mean)/(sqrt(2)*std)))*0.5)
mu_ = tic * mu
std_ = sqrt(tic) * std
print(round(norm(max_tic, mu_, std_), 4))
Day 6: The Central Limit Theorem III
\[\textrm{Confidence Interval} = \bar x \pm z {\sigma\over\sqrt n}\]# read inputs
sample = int(input())
mu = int(input())
std = int(input())
inte = float(input())
z = float(input())
# lower limit
print(round(mu - z*(std/(sample)**0.5), 2))
# higher limit
print(round(mu + z*(std/(sample)**0.5), 2))
Day 7: Pearson Correlation Coefficient I
\[\mu_X = {\sum_{i=1}^n x_i\over n}, similarly\;\mu_Y\] \[\sigma_X = \sqrt{\sum_{i=1}^n (x_i - \mu)^2\over n}, similarly\;\sigma_Y\] \[\rho_{X,Y} = {\Sigma(x_i-\mu_X)\cdot(y_i-\mu_Y)\over{n\cdot \sigma_X\cdot \sigma_Y}}\]# Read inputs
n = int(input())
X = list(map(float, input().split()))
Y = list(map(float, input().split()))
# check n == len(X) == len(Y)
# mean
def mean(a):
return sum(a)/len(a)
# standard deviation
def std(a):
global n
mu = mean(a)
sum_ = 0.0
for i in range(n):
sum_ += ((a[i] - mu)**2)
return (sum_/n)**0.5
# corr(X, Y)
def corr(arr1, arr2):
global n
mean1 = mean(arr1)
mean2 = mean(arr2)
conv = 0.0
for i in range(n):
conv += (arr1[i] - mean1) * (arr2[i] - mean2)
print(round(conv/(n*std(arr1)*std(arr2)), 3))
corr(X, Y)
Day 7: Spearman’s Rank Correlation Coefficient
\[r_x \textrm{: rank of X values in descending order}\] \[r_y \textrm{: rank of Y values in descending order}\] \[N \textrm{: number of values in X or Y, }N_X = N_Y\] \[1\leq r_x,r_y \leq N\] \[r_{xy} = 1-{6\Sigma(r_x - r_y)^2\over{N(N^2-1)}}\]# Read inputs
n = int(input())
X = list(map(float, input().split()))
Y = list(map(float, input().split()))
# Ranking
def rank(arr):
return [sorted(arr).index(x)+1 for x in arr]
# Spearman's rank correlation
def corr(arr1, arr2):
global n
r_arr1 = rank(arr1)
r_arr2 = rank(arr2)
sum_ = 0.0
for i in range(n):
sum_ += (r_arr1[i] - r_arr2[i])**2
print(round(1-((6*sum_)/(n*((n**2)-1))), 3))
corr(X, Y)
Day 8: Least Square Regression Line
\[a = {n\Sigma(x_iy_j)-(\Sigma x_i)(\Sigma y_j)\over{n\Sigma(x^2_i)-\Sigma(x_i)^2}}\] \[b = {\Sigma(x_i-\mu_X)\cdot(y_i-\mu_Y)\over{n\cdot \sigma_X^2}}\] \[\hat{Y} = a + bX\]# Read inputs into array
n = 5
X, Y = list(), list()
for _ in range(n):
x, y = map(int, input().split())
X.append(x)
Y.append(y)
# mean
def mean(a):
return sum(a)/len(a)
# standard deviation
def std(a):
global n
mu = mean(a)
sum_ = 0.0
for i in range(n):
sum_ += ((a[i] - mu)**2)
return (sum_/n)**0.5
# corr(X, Y)
def corr(arr1, arr2):
global n
mean1 = mean(arr1)
mean2 = mean(arr2)
conv = 0.0
for i in range(n):
conv += (arr1[i] - mean1) * (arr2[i] - mean2)
return conv/(n*std(arr1)*std(arr2))
# fit the best-fit line using least squares and find the value of y
def liner(X, Y, x1):
b = corr(X,Y)*(std(Y)/std(X))
a = mean(Y) - (b*mean(X))
print(round(a + (b*x1), 3))
liner(X, Y, 80)
Day 9: Multiple Linear Regression
Working on a bug. Will be posted soon.
If you encounter any error, feel free to post your issue in GitHub.
Ⓒ Surya Pusapati 2022