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A discrete random variable \(X\) can assume five possible values: 12,13,15,18 and 20\. Its probability distribution is shown below. $$\begin{array}{|l|c|c|c|c|c|} \hline x & 12 & 13 & 15 & 18 & 20 \\ \hline P(x) & 0.14 & 0.11 & & 0.26 & 0.23 \\\\\hline \end{array}$$ a) What is \(P(15) ?\) b) What is the probability that \(x\) equals 12 or \(20 ?\) c) What is \(P(x \leqslant 18) ?\) d) Find \(E(X)\) e) Find \(V(X)\) f) Let \(Y=0.5 X-4 .\) Find \(E(Y)\) and \(V(n)\) g) Compare your results in \(\mathrm{d}\) ), \(\mathrm{e}\) ) and \(\mathrm{f}\) ) and generalize for \(Y=a X+b,\) where \(a\) and \(b\) are constants.

Short Answer

Expert verified
a) 0.26; b) 0.37; c) 0.77; d) 16.29; e) 8.31; f) E(Y) = 4.145, V(Y) = 2.0775; g) E(Y) = aE(X)+b, V(Y) = a^2V(X).

Step by step solution

01

Find Missing Probability

To find \(P(15)\), recall that the probabilities must sum up to 1. Sum the given probabilities: \(0.14 + 0.11 + 0.26 + 0.23 = 0.74\). So, \(P(15) = 1-0.74 = 0.26\).
02

Calculate Probability for x = 12 or 20

To find the probability that \(x\) equals 12 or 20, simply add their probabilities: \(P(12) + P(20) = 0.14 + 0.23 = 0.37\).
03

Determine Cumulative Probability for x ≤ 18

Calculate \(P(x \leq 18)\) by adding the probabilities of all values \(\leq 18\): \(P(12) + P(13) + P(15) + P(18) = 0.14 + 0.11 + 0.26 + 0.26 = 0.77\).
04

Compute Expected Value E(X)

The expected value \(E(X)\) is calculated by summing each value multiplied by its probability: \(E(X) = 12\times0.14 + 13\times0.11 + 15\times0.26 + 18\times0.26 + 20\times0.23 = 1.68 + 1.43 + 3.9 + 4.68 + 4.6 = 16.29\).
05

Find Variance V(X)

First, find \(E(X^2)\): \(E(X^2) = 12^2\times0.14 + 13^2\times0.11 + 15^2\times0.26 + 18^2\times0.26 + 20^2\times0.23 = 20.16 + 18.59 + 58.5 + 84.24 + 92 = 273.49\). Use \(V(X) = E(X^2) - (E(X))^2\): \(V(X) = 273.49 - 16.29^2 = 273.49 - 265.1841 = 8.3059\approx8.31\).
06

Transform Random Variable and Find E(Y) and V(Y)

For \(Y = 0.5X - 4\), \(E(Y) = 0.5E(X) - 4 = 0.5 \times 16.29 - 4 = 8.145 - 4 = 4.145\). For variance \(V(Y) = (0.5)^2V(X) = 0.25\times8.31 = 2.0775\).
07

Generalize for Y = aX + b

The general formulas are \(E(Y) = aE(X) + b\) and \(V(Y) = a^2V(X)\). Thus, transforming by a constant \(b\) does not alter variance, whereas multiplying by \(a\) scales the variance by \(a^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value is a key concept in probability and statistics, representing the average or mean value of a random variable in a probabilistic context. To compute the expected value, or \(E(X)\), we sum the products of each possible value that the discrete random variable \(X\) can assume and its corresponding probability. Mathematically, it's expressed as:

\[E(X) = \sum_{i} x_i \times P(x_i)\]Where:
  • \(x_i\) are the possible values of the random variable \(X\)
  • \(P(x_i)\) are the probabilities of each value \(x_i\)
In the presented exercise, by using the given values and their probabilities, we calculated \(E(X) = 16.29\). This means that on average, the random variable \(X\) would be around 16.29 if the experiment were repeated many times.

Expected value is fundamental when analyzing random events, as it allows for predicting long-term behavior of the variable. Be sure to weigh all possible outcomes by their probabilities to get accurate results.
Variance
Variance gives us insight into the spread of a set of values, indicating how much the values of a random variable differ from the expected value. It's defined as the expected value of the squared deviations from the mean. Mathematically, variance \(V(X)\) is computed as:

\[V(X) = E(X^2) - (E(X))^2\]Where:
  • \(E(X^2)\) is the expected value of the square of \(X\).
  • \(E(X)\) has been previously determined.
For the discrete random variable in this case, \(V(X) \approx 8.31\). This tells us that the values of \(X\) are, on average, around 8.31 units away from the mean value \(16.29\).

Variance is crucial for understanding data variability. It allows statisticians and experimenters to assess the reliability and consistency of datasets. A higher variance means more variability in data points, whereas a lower variance implies that data points are more clustered around the mean.
Discrete Random Variable
A discrete random variable is a type of random variable that can take on a finite number of distinct outcomes. Unlike continuous random variables which can assume any value within a range, discrete ones, such as rolling a die or flipping a coin, have countable possible outcomes.

In probability, understanding discrete random variables involves knowing their distributions, which are summarized by their probabilities. The sum of all these probabilities equals 1. In the given problem, the random variable \(X\) can take values 12, 13, 15, 18, and 20, with a defined probability for each.

To work effectively with discrete random variables:
  • Always ensure the sum of probabilities equals 1.
  • Know how to compute expected values and variances to understand the average behavior and spread.
  • Familiarize yourself with transformations, such as \(Y = aX + b\), to predict how mathematical operations affect the variable's behaviour.
Discrete random variables are foundational in statistics, providing a straightforward yet powerful tool to model real-world scenarios in discrete contexts.

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