Basics of ML & AL Week 2 Challenge
1.Ans
Calculating the four business moments:
A)Calculating the Mean of the problem:
P(i)*X(i)
= (0*0.35)+ (1*0.25)+(2*0.15)+(3*0.15) + (4*0.10)
= 0+ 0.25 +0.3+0.45+0.4 =1.4
B). Variance
Sum(Xi-Mu)^2 *P(X)
= (0-1.4)^2 *0.35 +((1-1.4)^2)*0.25 + ((2-1.4)^2)*0.15 + ((3-1.4)^2)*0.15 + ((4-1.4)^2)*0.10
= 0.686 + 0.04 + 0.054 + 0.384 + 0.676 = 0.78
Variance = 1.84
SD = 1.35
C) Skewness
= ((0-1.4/1.35)^3 *0.35 +((1-1.4/1.35)^3)*0.25 + ((2-1.4/1.35)^3)*0.15 + ((3-1.4/1.35)^3)*0.15 + ((4-1.4/1.35)^3)*0.10
= -0.38 - 0.0069 + 0.013+ 0.244 + 0.6987
= 0.569
D) Kurtosis
Sum[((Xi-Mu)/SD)^4 * P(X)] - 3
= [((0-1.4/1.35)^4 *0.35] +[((1-1.4/1.35)^4)*0.25] + [((2-1.4/1.35)^4)*0.15] + [((3-1.4/1.35)^4)*0.15] + [((4-1.4/1.35)^4)*0.10] - 3
= [0.39 +0.0018 + 0.0056 +0.28 + 1.335] - 3
= 2.0124 - 3
= - 0.98
2. The significance of expected value is important when the propability factor is involved. Without the probability of an occurence the mean i.e the sum of observations / the number of observations will be enough.
For ex. if a sample is taken when patients are likely to expire in covid in a day
age factor probability to expire
0-18 0.05
18-25 0.1
25-45 0.15
45-60 0.3
60 and above 0.4
Each occurence if defined by the probability factor of the occurence. So for this reason PiXi the expected value is important.
3.Skewness is a very important phenomena which helps in suggesting how the distribution is along the curve. It suggests how the various points are scattered and how the Mean, median and mode is projected. No skewness is an absolute normal distribution with symmetric values spread throughout which is unlikely to be possible in most of the cases.
4. Evaluating the probabilities of the normal distribution with Mu(mean) =680
SD = 31
Z = (X - Mu)/ SD
a) P(X<711)
Z = 711 - 680/31 = 1
P(Z = 1) = 0.8413
84%
b) P (X>740)
Z = 740 - 680 / 31 = 1.93
P(X>740) = 1-P(1.93) =1 -0.9732
= 0.0268
2%
c))P(600<X<720)
Z(600) = 600 - 680 / 31 = -2.58
Z (720) = 720 -680 /31 = 1.29
Now P(higher value - lower value)
P(1.29)-P(2.58) = 0.9015 - 0.0494
= 0.8521
85%
d) P(X=720)
Z = 720 - 680 / 31 = 1.29
P(1.29) = 0.9015
90% which makes this the ideal standard distribution
5. This normal distribution graph shows the symmetricity of the data where the mean comes in the center and one sigma is 34.1+34.1 = 68.2%.
And 2 sigma = 34.1+34.1+ 13.6+13.6 = 95.4. So for 2 sigma almost 95.4% of the values fall into the distribution graph
Similarly for 3 sigma it is taken as 34.1+34.1+13.6+13.6+2.1+2.1 = 99.6% falls in the 3 sigma range