Basics of ML & AL Week 2 Challenge

1)

2)

When the random values are finite and small then the mean can be calculated directly by using the

simple mean = sum of observations /number of observations

when the number of observations are infinite or large number then the calculation of each and every outcome will be hard.In such cases, the probability of each outcome is determined using a sample study and using this probability the expected value as following:

μ=∑x.P(x)

3)

Skewness is a data characteristic which measures the symmetricity of the curve based on the data. 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, where in negative skewness, the mean value is higher than the median and mode.

4)

P(X<711)

P(X<711) = P(680 + 31) = P(μ+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 1σ

is 0.34

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=(740−680)/31=1.935=1.94

P(Z<1.94) = 0.97381

c)P(600<x<720)< p=""></x<720)<>

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

P(X=720) = 0

5)

The curve that is shown is a nomal curve since it symmetric in nature. Here mean, median and mode lies in the centre of the curve which divides the curve symmetrically.The area under the curve between mean and first standard deviation is 34.1% where it indiactes that the out come of an event has 34.1% chance to fall under this area.It is also can be observed that the area between μ+1σandμ+2σ is 13.6% of the total area and μ+2σandμ+3σ is 2.1% and outliers as 0.1%.

So in total the area of the curve between μ+1σ and μ+3σ on right side is 49.8% out of 50%. The mean value is apporximately 0.4 so the mean median and mode can be assumed as same with value of 0.4