Basics of ML & AL Week 2 Challenge

1)

2)

Expected value is the average value of random variable and it is an important summary value of the distribution of the variable. The mean value will take the all observations which will suit for small data or population and we cannot take all the observations sometimes,so at that moment the excepted value will give the result for analysis.

3)

Because of the deviation of the data from the original distribution , it is considered bad.

4)

Given data \mu = 680 , \sigma = 31

a) P(X<711):

`Z = \frac{X - \mu }{ \sigma }`

= (711-680)/31 = 31/31 = 1

From Z table for the standard score of 1 and the mean of 0.

P(X<711) = 0.84134 = 84.31%

b) P(X>740)

We know the prob any event is 1 , P(X>740) = 1 - P(X<740)

`Z = \frac{X - \mu }{ \sigma }`

= (740-680)/31 = 60/31 = 1.93

Z = 1.9

From Z table for the standard score of 1.9 and the mean of 0.

P(X<740) = 1 -0.97128 = 0.0287

P(X<740) = 2.87%

c) P(600<X<720)

For prob of X<720:

`Z = \frac{X - \mu }{ \sigma }`

= (720-680)/31 = 40/31 = 1.29

Z ~= 1.3

From Z table for the standard score of 1.3 and the mean of 0 for P(X<720) = 0.90320

For P(X<600)

= (600-680)/31 = -80/31 = -2.58

Z = -2.6

From Z table for the standard score of 1.3 and the mean of 0 P(600<X) = 0.00466

for P(600<X<720) = 0.90320 - 0.89851

P(600<X<720) = 89.8%

d) P(X=720) = 0

For continuous variable probability at any point is zero

5)

The bell curve portrayed in the question is probability distribution function for the normal distribution

In a normal distribution , it decribes that the data is Mean centric i.e; 50% of the data above mean the remaining 50% below

the mean

`\sigma` being the standard deviation

In general 64.2% data lies between - `-\sigma` to + `\sigma`

In general 95.4% data lies between -`-2\sigma` to +2`\sigma`

In general 99.7% data lies between - `-3\sigma` to +3`\sigma`

In general 99.999% data lies between - `-6\sigma` to +6`\sigma`