Basics of ML & AL Week 2 Challenge
1ans) Bussiness Moment - Mean
μ= ∑( X . 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 + 0.45 + 0.4
μ= 1.4
Bussiness Moment - variance
σ^2 = ∑ ( X - μ )^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
= 1.84
= 1.36
Bussiness moment - skewness
= ∑ (x-μ/σ)^3 . P(X)
= ( 0- 0.14 /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
= (- 0.38) + (-0.0060) + 0.0129 + 0.2442 + 0.6987
Skewness = 0.57
Bussiness moment - kurtosis
= ∑ ( X - μ/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
0.393+0.00187+ 0.00568+0.287 + 1.3358 -3
= -0.98
2 Ans ) An Expected value gives a quick insight into the behaviour of a random variable without knowing if it is discrete or continuous.
Therefore , two variables with the same expected value can have different probability distribution . Other descriptive measures like standard deviation also affect the shape of the distribution. The expected value of random variable X is denoted as E(X).
3Ans) Skewness refers to lack of symmetry in a symmetrical bell curve , or normal distribution in a set of data. If the curve is shifted to the left or right , it is said to be skewed . Skewness can be qualifed as a represention of the extent to which a given distribution varies from a normal distribution . A normal distribution of zero.
Having skewness in the curve is not a bad analysis, the skewness tells about the curve is symmetric or not and it also helps us to plot the different curves in the single one . Where the median and mode will be the same for different curves and thus tells us whether the curve is symmetric or not symmetric.
Mean : Averange of the data sets
Median : Central value in the data set
Mode : Repeating number in the data set
4ans )
Z= X- μ/σ
Where X = Random variable
Mean (μ) = 680
Standard deviation = 31
A) P(X<711)
Z = 711- 680/31
Z= 1
P( Z<1) = 0.8413 from the Z table
Area to the left of the Z score scince X is the less than 711
B) P( X>740)
Z= 740- 680/31
Z= 1.94
P(z>1.94) = 0.9738
the area to the left of the Z score ,
but we need to calculate the right side area also
P(Z>1.94) = 1- P(Z<1.94) = 1- 0.9732
= 0.0268
C) P(600<x>720)
Z= 600 - 711/31 and 720 -711/31
Z = - 2.58 and
Z = 1.29
we need to find the area of both z score -2.58 and 1.29 subract the values to get the probabilty
P(-2.58 < 1.29) = P(<1.29) - P(Z<-2.58)
= 0.9032-0.0047
= 0.89854
D) P(X=720)
Z= 1.29
P(Z= 1.29) = 0.9015
= 0.9015 - 0.9015
= 0
5Ans) That the given curve is Normal Distribution Curve for the random variable and the Percentage is the 'Standard Deviation ' from the Mean. The distribution in the curve is perfect without any skewness.
The normal distribution is a Probabilty function that describes the value of avariable are distributed.
σ ( standard Deviation) from the mean is around 34.1 on the both sides of the mean and the values lies between -34.1% and +34.1% = 68.2%of the values consider as the first standard deviation region.
2σ from the mean is taken as 34.1%+34.1%+13.6%+13.6%= 95.4% approximately values is consider as Second standard deviation from the Mean.
3σ from the is taken as 34.1%+34.1%+13.6%+13.6%+2.1%+2.1%= 99.7 approximately the value consider as a third standard devition from the mean.