Basics of ML & AL Week 2 Challenge

1.Calculate all 4 business moments using pen and paper for the below data set?

Mean `mu` =∑ X x P(X)

= (0x0.35) +(1 x0.25) +(2 x0.15) +(3x0.15) +(4x0.10)

=1.4

Variance `sigma^2`=`sum(X_i-mu_x)^2 X P(x)`

=(0-0)² X(0.35) +(1-0.25)² X(0.25) +(2-0.3)² X(0.15) +(3-0.45)² X(0.15) +(4-0.4)²X(0.1)

=0+0.140+0.4335+0.9753+1.296

= 2.8448

Standard deviation `sigma` =`sqrtsigma^2`

= `sqrt2.8448` = 1.686

Skewness =`sum((X_i-mu_x)/sigma^3)X P(X)`

= ((0-0)/1.686)³ X 0.35 +((1-0.25)/1.686)³ X0.25 +((2-0.3)/1.686)³ X0.15 +((3-0.45)/1.686)³ X0.15 +((4-0.4)/1.686)³ X 0.10

=0+ 0.021+0.153+0.518+0.973

=1.665

Kurtosis =`sum((X_i-mu_x)/sigma^4)X P(X)-3`

= ((0-0)/1.686)⁴ X(0.35)-3 +((1-0.25)/1.686)⁴ X(0.25)-3 +((2-0.3)/1.686)⁴ X(0.15)-3 +((3-0.45)/1.686)⁴ X(0.15)-3 +((4-0.4)/1.686)⁴ X (0.10)-3

= -3 +(0.009-3) +(0.155-3) +(0.784-3)+(2.07-3)

=-11.978

3. Having skewness in the curve considered to be bad in the analysis?

Yes in the dataset we have skewness is present then we conclude that not evenly distributed. And We said that most range of values is present either left or right side. And so many outliers are formed we taken that skewness range. That outliers will effect our data prediction also.

And negative skewness cause mean and median less than mode. And Positive skewness cause mean, median greater than mode. For Example We taken a Cancer Patients Hospital Dataset Then Most of people are lived at 2^nd stage only(80%) small people lived at fourth stage(5%) that means it skewed left we taken distribution and lot of outliers formed at fourth stage. Then we cannot predict well. Skewness not bad for Analysis. But Bad For Prediction\

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)

Given That

Mean µ = 680

Standard deviation = 31

1. P(X<711)

Z score = (X-`mu`)/`sigma` )

= (711-680)/31 = 1.00 = 0.84134 = 84.134%

2.P(X>740)

= (740-680)/31 =1.935 1.9 under 3 1-P (1-0.97320) = 0.0268 or 0.0264 = 2.68%

3.P(600<X<720)

P(600) = (600-680)/31 = -2.58 -2.5 under 8 =0.00494

P(720) = (720-680)/31 = 1.290 1.29 under 0 = 0.90147

89.653%

4.P(X= 720)

P(720) = (720-680)/31= 1.290 1.2 under 9

=0.90147 = 90.147% or 1- 0.29 = 0.29 = 0.11409 = 11.40%.

Then remove right side also.

Then it is negligible =0.

2.What is the significance of expected value when simple mean (Sum of all observations/number of observations) is already in place.

There is no difference between mean and the Expected value. But The highest and very lowest values will effect on the output then mis-understood the data and then think and take decision incorrectly. But Expected Value is calculated by individually calculated and the Probability of each value taken in to consideration. It gives correctly.

We take the First question, Example Mean is 2. And Expected value is 1.44. 0 and the 5 influence more on mean. So Expected value is more important to taken into consideration.

5.Explain the curve on the right side

Here The Total 68.2% data Fall under the -1 to 1. And the 95.4% data fall between -2 to 2. The X- Axis is taken Z- Score and the frequency is taken on Y- axis. It is normal distribution curve. For this one we take riding of bicycle then first he goes slow then he increases speed. While his legs are pain then he reduces the speed of Bicycle Riding.

And In this curve 0 to 1 are 34.1% data are there and the 47.7% data under the 0-2 and only 2 are 13.6% are there and symmetric on left side also.

And job opportunities from Nov – May

Peaks in Jan, Feb, Mar.
Nov, Dec some-what and April, May Some-what.