Basics of ML & AL Week 2 Challenge

Answer 1:

Random Variable (xi)	P(Xi)	xi*P(xi)	xi-mean	(xi-mean)^2	P(Xi)*(xi-mean)^2	(xi-mean)/SD	((xi-mean)/SD)^3*P(X)	((xi-mean)/SD)^4*P(Xi)
0	0.35	0	-1.4	1.96	0.686	-1.032	-0.385	0.00767308
1	0.25	0.25	-0.4	0.16	0.04	-0.295	-0.006	4.22195E-10
2	0.15	0.3	0.6	0.36	0.054	0.442	0.013	4.25956E-09
3	0.15	0.45	1.6	2.56	0.384	1.180	0.246	0.000550794
4	0.1	0.4	2.6	6.76	0.676	1.917	0.704	0.024590833

Mean	1.4
Variance	1.84
Standard Deviation	1.356465997
Skewness	0.572138895
Kurtosis	-2.967185289

Answer 2:

Let us assume that we have the number of outcomes `n_i` for every random variable `f_i`.

If we look at the formula for calculating simple mean of a random variable if we have the above data then we get the following:

`(f_1*n_1 + f_2*n_2 + ... +f_N*n_N)/(sum n_i)`

The above formula could be written as follows:

`f_1*P_1 + f_2*P_2 + ... +f_N*P_N`

Where,

`P_i = n_i/ (sum n_i)` is nothing but probability of occurrence of a given random variable.

Which is also the formula of the expected value of a random variable. So they both are the same.

Answer 3:

Skewed data means that the data we are dealing with is having a shape different from the symmetrical shape of a normal distribution. If the mode is lesser than mean and median then the data is positively skewed and if the mode is greater than mean and median then the data is negatively skewed.

Skewed data is difficult to handle.

Answer 4

Converting to the standard normal variate and using the normal distribution table we get the following results:

(a) P(X<711) =0.841

(b) P(X>740) =0.0268

(c) P(600<X<720) =0.8966

(d) P(x=720) =0 (Probability of occurence of a single point is practically zero in a continuous distribution)

Answer 5

The curve shown on the right side is a normal distribution curve. It tells us that 68.2% of data points lie between +`sigma` to -`sigma` and 47.7X2 i.e. 95.4% of data points lie within+2`sigma` to -2`sigma`.