Basics of ML & AL Week 2 Challenge

Ans. 1: Question one is solved on paper and the file is attached.

Ans. 2: Expected value is used to calculate the average of a population when the population size is infinite or very large and it is not possible calculate it . Instead of population size ,when frequency of the events are known then expected value is used to calculate average.

Ans. 3: Skewness is considered bad because, if there are too much skewness in data, then many statical model do not work. The tail region act as an outlier for the model and we know that outlier adversely affects the model performance.The mean,mode and median may be wrongly anticipated.

So, skewed data is transformed to either Gaussian distribution or normal distribution for analysis.

Ans. 4:

(a) P(x<711)= 0.5+(0.6826/2)

P(x<711)=0.8413

= 84.13%

(b) P(x>740)

The z score corresponding to x=740, with mean of 680 and standard deviation=31 is given by:-

Z=(x- μ)/σ

=(740-680)/31

=1.94

Now, calculating the value of probability from z table corresponding to the value of 1.94

P(z<1.94) =0.9738 (from z table)

P(x>740) = 1 - P(z<1.94) = 1 - 0.9738 = 0.0262

=2.62%

(c) P(600<x<720)

The z score corresponding to x=600 and x=720

Z(x=720) = (x- μ)/σ =(720-680)/31

= 1.29

Z(x=600) = (x- μ)/σ =(600-680)/31

= -2.58

P(z<1.29) = 0.9015

P(z<-2.58) = 0.0049

so, P(600<x<740) = P(z<1.29) - P(z<-2.58)

= 0.9015 - 0.0049

= 0.8966

= 89.66%

(d) P(x=720) = 0

In countinuous normal distribution the probability at any particular value of x is zero.

Ans. 5: The curve is a standard normal distribution with mean(μ)=0 and standard deviation(σ)=1.

68.2% of the values lies between 1 standard deviation from mean.

95.44% of the values lies between 2 standard deviation from mean.

99.74% of the values lies between 3 standard deviation from mean.

Any normal distribution can be converted to standard normal distribution by converting its values to Z-scores. Z-scores tells you how many standard deviations from the mean each value lies.