Basics of ML & AL Week 2 Challenge
1.Calculate all 4 business moments using pen and paper for the below data set?
1) mean µ= ∑(x1) * p(x)
= 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
2) Second business movement
Variance σ2 = ∑(x1-µx)^2 *P(x)
σ2 = (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
σ2 = 0.686+0.04+0.054+0.384+0.676
σ2 = 1.84
3) Third Business movement
σ = 1.356
Skewness (SK) =∑((xi-µx)/σ)^3* P(x)
= ((0-1.4)/1.356)^3 *0.35 + ((1-1.4)/1.356)^3 *0.25 + ((2-1.4)/1.356)^3 *0.15 + ((3-1.4)/1.356)^3 *0.15 + ((4-1.4)/1.356)^3 *0.10
= -0.3851+ (-0.00641) + 0.0129 + 0.246 + 0.704
= 0.57139
4) Forth Business Movement
Kurtosis (Kt) =∑((xi-µx)/σ)^4 *p(x) -3
= (((0-1.4)/1.356)^4*0.35 + ((1-1.4)/1.356)^4*0.25+ ((2-1.4)/1.356)^4*0.15+ ((3- 1.4)/1.356)^4*0.15 + ((4-1.4)/1.356)^4*0.10)-3
= (0.3976 + 0.00189 + 0.00574 + 0.2907 + 1.3516)-3
= -0.97665
= -0.98
2.What is the significance of expected value when simple mean (Sum of all observations/number of observations) is already in place
The applicablity of simple mean and expected value are dependent on the nature of the case study.
In smaller case studies where the random variable outcomes can be measured as frequency of occurences or each observation in the sample can be noted, the simple mean is calculated as Sum of all observations/number of observations.
However, in a case of an infinite number of observations, it is very difficult to note each and every outcome of the given data set. In such cases, the probability of each outcome is determined using a sample study and using this probability the expected value is determined using:
Mean or expected value, μ=∑x.P(x)μ=∑x.P(x)
The expected values reduces the analysis expenditure and serves the purpose of the measuring the central tendency of the data set
3.Having skewness in the curve considered to be bad in the analysis?
Skewness is a data characteristic which measures the symmetricity of the curve. It is neither good nor bad for the analysis.
For a normal distribution, the mean, median and mode coincide. However, this is an ideal case and most of the real data have either postive or negative skewness.
A positive skewness indicates that the median and mode are greater than the mean value. On the other hand, for a negatve skewness, the mean value is higher than the median and mode.
4.Evaluate the probabilities for continuous normal distribution with given mean = 680 and standard deviation = 31
a)P(X<711)
b)P(X>740)
c)P(600<x<720)< p=""></x<720)<>
d)P(X=720)
a)P(X<711)
P(X<711) = P(680 + 31) = P(μ+1σμ+1σ)
In a normal distribution, the area under the curve on the left, which is the probability of the outcomes, is 0.5
Also, area under the curve from origin to 1σσ is 0.34
Therefore, P(X<711) = 0.5 + 0.34 = 0.84
b)P(X>740)
Value of Z corresponding to X=740,
Z=(X−μ)/σ Z
Z=(740−680)/31=1.935=1.94Z=740-68031=1.935=1.94
Using the Z-table, the probability P(Z<1.94) = 0.97381
c)P(600<x<720)< p=""></x<720)<>
This can be rewritten as P(-2.58<Z<1.29)
P(Z<-2.58) = 0.00494
P(Z<1.29) = 0.90147
P(-2.58<Z<1.29) = P(Z<1.29) - P(Z<-2.58) = 0.90147 - 0.00494 = 0.897
P(600<x<720) =0.897
d)P(X=720)
Z(X=720) = 1.29
For calculating P(Z=1.29), we need to subtract the area on the left as well the right of the curve from the total area.
P(Z=1.29) = 1 - P(Z<1.29) - P(Z>1.29) = 1 - 0.90147 - (1-0.90147) = 0
Therefore, P(X=720) = 0
5.Explain the curve on the right side
The curve shown in the figure represents a normal distribution. The peak of the curve divides the curve and the area is symmetric about the y-axis with the mean or expected value located at the origin. The right side of the curve is divided using the standard deviation as follows:
The area under the curve between the mean and 1 standard deviation is 34.1% of the total area. This means that 34.1% of the outcomes lie in this region.
Also, the area between μ+1σandμ+2σμ+1σandμ+2σ is 13.6% of the total area