Basics of ML & AL Week 2 Challenge
Q2
The expected value and the simple mean are related concepts but they serve different purposes in machine learning.
The simple mean is a basic statistical measure that represents the average value of a set of observations. It is calculated by adding up all the values in the set and dividing by the number of observations. The simple mean provides a quick and easy way to summarize a data set, but it doesn't take into account the likelihood of each value occurring.
On the other hand, the expected value is a concept from probability theory that takes into account the likelihood of each possible value occurring. It is a weighted average of all possible outcomes, where the weights are the probabilities of each outcome. In machine learning, the expected value can be used to estimate the outcome of a random variable, given a set of possible outcomes and their corresponding probabilities.
The expected value is particularly useful in situations where the outcome is uncertain, but the probabilities of each possible outcome are known. For example, in decision theory, the expected value can be used to determine the best course of action when faced with different options and their associated probabilities.
In summary, the expected value is a more sophisticated measure than the simple mean because it takes into account the likelihood of each value occurring. In machine learning, it can be used to estimate the outcome of a random variable when the probabilities of each possible outcome are known.
Having skewness in a curve is not necessarily bad in machine learning analysis, but it can affect the performance of some models and the interpretation of the results.
Skewness is a measure of the asymmetry of a probability distribution, and it can be positive (right-skewed), negative (left-skewed), or zero (symmetric). In some cases, a certain degree of skewness may be expected or even desired, such as when the data represents a real-world phenomenon that is naturally skewed. For example, the distribution of income is typically right-skewed, with a few high earners skewing the distribution towards the right.
However, in other cases, skewness may indicate issues with the data or the model. For example, if the response variable in a regression model is heavily skewed, it may be more difficult to accurately model the relationship between the predictor variables and the response variable. Skewed data may also violate the assumptions of some statistical tests, leading to biased or incorrect results.
In summary, having skewness in a curve is not inherently bad in machine learning analysis, but it can affect the performance of some models and the interpretation of the results. It is important to understand the causes of the skewness and to take appropriate measures, such as transforming the data or using models that are robust to skewness, to ensure that the analysis is accurate and reliable.
The continuous normal distribution with mean $\mu$ and standard deviation $\sigma$ is given by:
�(�)=1�2��−(�−�)22�2f(x)=σ2π1e−2σ2(x−μ)2
where $x$ is any value in the domain of the distribution.
a) P(X<711)
To find this probability, we need to find the area under the normal distribution curve to the left of 711.
�(�<711)=∫−∞711�(�)��=Φ(711−��)P(X<711)=∫−∞711f(x)dx=Φ(σ711−μ)
where $\Phi$ is the cumulative distribution function of the standard normal distribution, evaluated at the standard score $\frac{711-\mu}{\sigma}$.
Substituting the given values, we get:
�(�<711)=Φ(711−68031)=Φ(1)=0.8413P(X<711)=Φ(31711−680)=Φ(1)=0.8413
Therefore, the probability that $X$ is less than 711 is 0.8413.
b) P(X>740)
To find this probability, we need to find the area under the normal distribution curve to the right of 740.
�(�>740)=∫740∞�(�)��=1−Φ(740−��)P(X>740)=∫740∞f(x)dx=1−Φ(σ740−μ)
Substituting the given values, we get:
�(�>740)=1−Φ(740−68031)=1−Φ(1.9355)=1−0.9749=0.0251P(X>740)=1−Φ(31740−680)=1−Φ(1.9355)=1−0.9749=0.0251
Therefore, the probability that $X$ is greater than 740 is 0.0251.
c) P(600
To find this probability, we need to find the area under the normal distribution curve between 600 and 720.
�(600<�<720)=∫600720�(�)��=Φ(720−��)−Φ(600−��)P(600<X<720)=∫600720f(x)dx=Φ(σ720−μ)−Φ(σ600−μ)
Substituting the given values, we get:
�(600<�<720)=Φ(720−68031)−Φ(600−68031)=Φ(1.2903)−Φ(−2.5806)P(600<X<720)=Φ(31720−680)−Φ(31600−680)=Φ(1.2903)−Φ(−2.5806)
Using the table of the standard normal distribution or a calculator, we get:
�(600<�<720)=0.9015−0.0052=0.8963P(600<X<720)=0.9015−0.0052=0.8963
Therefore, the probability that $X$ is between 600 and 720 is 0.8963.
d) P(X=720)
The probability that a continuous random variable takes on a specific value is always zero. Therefore, $P(X=720) = 0$.
Q4