Basics of ML & AL Week 2 Challenge
Question 1:
1] First Business Moments:-Measure of central Tendancy
Mean(u) = ∑((xi*P(x))
=[(0 x 0.35) + (1 x 0.25) + (2 x 0.15) + (3 x 0.15) + (4 x 0.10) ]
= 0 + 0.25 + 0.30 + 0.35 + 0.4
=1.4
2]Second Business Moments:- Measure of dispersion
Variance(σ^2) = ∑(((xi-ux)^2 * P(x))
=[((0-1.4)^2 x 0.35) + ((1-1.4)^2 x 0.25) + ((2-1.4)^2 x 0.15) + ((3-1.4)^2 x 0.15) +((4-1.4)^2 x 0.10) ]
= 0.686 + 0.04 + 0.054 + 0.384 + 0.676
σ^2 =1.84
σ = 1.36
3] Third Business Moments:- Skewness
sk = ∑( ( ( (xi-ux)/σ)^3 * P(x) )
= [ ( ( (0-1.4)/1.36)^3 x 0.35) + ( ( (1-1.4)/1.36)^3 x 0.25) + (( (2-1.4)/1.36) ^3 x 0.15) + ( ( (3-1.4)/1.36)^3 x 0.15)
+ ( ( (4-1.4)/1.36)^3 x 0.10) ]
= 0.568
4] Third Business Moments:- kurtosis
kt = ∑( ( ( (xi-ux)/σ)^4 * P(x) ) - 3
= [ ( ( (0-1.4)/1.36)^4 x 0.35) + ( ( (1-1.4)/1.36)^4 x 0.25) + (( (2-1.4)/1.36) ^4 x 0.15) + ( ( (3-1.4)/1.36)^4 x 0.15) + ( ( (4-1.4)/1.36)^4 x 0.10) ] - 3
= - 0.9762
.png file of paper calculation is also attached as mentioned in the question.
Question 2:
Expected value is long term average value of a random variable
Expected value isn't the precise value that you will definitely get this as the average for a large experiment. Rather it is a theoretical value denoting that the actual answer will be close to this value.
1st roll = (1+1+1+1+1+1+1)/6 =1
2nd roll = (1+2+3+1+4+5+2)/6 = 3
nth roll = (3+4+5+4+5+3)/6 =4
If we take average of first 2 averages= (1+3)/2 =2
if we take average of all the n sets, it would move closer to 3.5
Can we say it is expected to close down to value of 3.5, as n becomes very large.
Question 3:
Not neccassrily and very hard to determine. Different types of skewness depicts the data according to the domain of its application.
Question 4:
Z-score = (X – mean) / standard deviation.
a. P(X<711)
z-score = (711-680)/31 = 1.
z = +1 => 0.8413 => 84.13%.
b. P(X>740)
z-score = (740-680)/31 = 1.93.
z = +1.93 => 0.9732.
Probability greater than 740 => 1 - 0.9732 = 0.0268 => 2.68%.
c. P(600<x<720)
X = 720, z = 1.29 => 0.9015
X = 600, z = -2.58 => 0.00494
Probability value = 0.9015 – 0.00494 = 0.89656 => 89.656%.
d. P(X=720)
P(Z = 1.29) = P(1.29 < Z < 1.29)
Being a single line the area enclosed by it is Zero.
P(Z = 1.29) = 0.
Question 5:
The graph given in the question shows the normal distribution of the data points by plotting the z-values with their corresponding probability values.
The mean of this observation data is 0 and the standard deviation is 1.
It gives us the percentage of data falling under every 1, 2 and 3 standard deviation from the mean value.
In this figure, 68.2% of data falls from z values -1 to +1, 95.4% from -2 to +2 and 99.6% from -3 to +3.
**Week 2 Challenge Completed**
Thank you!!