Basics of ML & AL Week 2 Challenge

Ans 1)

Mean (μ): μ=∑ixi ⋅ P(xi)

μ=0⋅0.35 + 1⋅0.25 + 2⋅0.15 + 3⋅0.15 + 4⋅0.10

μ=0 + 0.25 + 0.30 + 0.45 + 0.40

μ=1.40

Variance (σ²): σ² =∑i(xi−μ)² ⋅ P(xi)

σ² =(0−1.40)²⋅0.35 + (1−1.40)²⋅0.25 + (2−1.40)²⋅0.15 + (3−1.40)²⋅0.15 + (4−1.40)²⋅0.10

σ² =1.84

σ= 1.36

Skewness (sk): sk= ∑i[((xi−μ)/σ)³⋅ P(xi)]

sk = [((0−1.40)/1.36)³⋅0.35] + [((1−1.40)/1.36)³⋅0.25] + [((2−1.40)/1.36)³⋅0.15] + [((3−1.40)/1.36)³⋅0.15] + [((4−1.40)/1.36)³⋅0.10)]

sk=0.568

Kurtosis (k): k= ∑i[((xi−μ)/ σ)^4 ⋅ P(xi)-3]

k= {[((0−1.40)/1.36)^4⋅0.35] + [((1−1.40)/1.36)^4⋅0.25] + [((2−1.40)/1.36)^4⋅0.15] + [((3−1.40)/1.36)^4⋅0.15] + [((4−1.40)/1.36)^4⋅0.10)]} - 3

k= -0.976

(Kindly see attachment for pen written solution)

Ans 2)

The expected value and the simple mean are related concepts, but they are not always the same, especially when dealing with random variables.

The simple mean (or arithmetic mean) is calculated by summing up all observations and dividing by the number of observations. It is a measure of central tendency that provides the average value of a set of numbers.

On the other hand, the expected value is a concept from probability theory. It is used to represent the average value of a random variable, taking into account the probabilities of different outcomes. For a discrete random variable, the expected value is calculated by summing the product of each possible value and its probability. For a continuous random variable, it involves integration.

Ans 3)

Skewness is a measure of the asymmetry of a probability distribution. In a perfectly symmetrical distribution, the left and right sides of the distribution are mirror images of each other, and the skewness is zero. If the distribution has a tail that extends more to one side than the other, it is considered skewed. Whether skewness is considered "bad" or not depends on the context and the goals of the analysis.

It's essential to interpret skewness in conjunction with other statistical measures and consider the implications for the particular field or application.

Ans 4)

The formula for this transformation is: Z= (X−μ)/σ​

a) P(X<711)= (711 − 680)/31 ≈ 1

Now, we look up the probability for Z=1 in the standard normal distribution table. From the table, P(Z<1)≈0.8413.

So, P(X<711)≈0.8413.

b) P(X>740)= (740 − 680)/31 ≈ 1.935

Now, we look up the probability for Z=1.935 in the standard normal distribution table. From the table, P(Z>1.935)≈0.026.

So, P(X>740)≈0.026.

c) P(600<X<720)= Z lower = (600 − 680)/31 ≈ − 2.581 Z upper = (720 − 680)/31 ≈ 1.290

Now, we look up the probabilities for Z lower ​ and Z upper ​ in the standard normal distribution table.

From the table: P(Z<−2.581)≈0.005 P(Z<1.290)≈0.901

So, P(600<X<720)= P(Z upper ​ )−P(Z lower ​ )≈0.901−0.005 =0.896.

d) P(X=720) For a continuous distribution, the probability of a specific point is zero. Therefore, P(X=720)=0.

Ans 5)

Here, the mean of this observation data is always 0 and the standard deviation is 1. It gives us the percentage of data falling under every 1, 2 and 3 standard deviation from the mean value. In this figure, 68.2% of data falls from z values -1 to +1, 95.4% from -2 to +2 and 99.6% from -3 to +3. An example application of this graph would be the results of IQ test of people living in a city. It can be noted that not everyone would have gotten the highest IQ or the lowest. Most of the range would be around the mean value which is depicted by this graph. Another example would be marks obtained in a test. It can be observed that people who score centum and people who scored a zero would be very less. More students’ marks would be around the average class mark.