/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 \(X\) has probability distributi... [FREE SOLUTION] | 91Ó°ÊÓ

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\(X\) has probability distribution as shown in the table. $$\begin{array}{|l|c|c|c|c|c|} \hline x & 5 & 10 & 15 & 20 & 25 \\ \hline P(x) & \frac{3}{20} & \frac{7}{30} & k & \frac{3}{10} & \frac{13}{60} \\\ \hline \end{array}$$ a) Find the value of \(k\) b) Find \(P(x>10)\) c) Find \(P(5

Short Answer

Expert verified
k is \(\frac{1}{10}\); \(P(x>10) = \frac{41}{60}\); \(P(5<x \leqslant 20) = \frac{7}{10}\); E(X) = 15.08; SD(X) \approx 6.88; E(Y) \approx 2.016; V(Y) \approx 1.896.

Step by step solution

01

Find the Sum of Probabilities

A probability distribution sums to 1. Therefore, to find the value of \(k\), we set up the equation: \[ \frac{3}{20} + \frac{7}{30} + k + \frac{3}{10} + \frac{13}{60} = 1. \] We find a common denominator for the fractions, which is 60. The equation becomes \[ \frac{9}{60} + \frac{14}{60} + k + \frac{18}{60} + \frac{13}{60} = 1. \] Simplifying this gives \(\frac{54}{60} + k = 1\), so \(k = 1 - \frac{54}{60} = \frac{6}{60} = \frac{1}{10}\).
02

Find P(x > 10)

To find \(P(x > 10)\), we sum the probabilities of all \(x\) values greater than 10: \[P(x > 10) = P(15) + P(20) + P(25) = \frac{1}{10} + \frac{3}{10} + \frac{13}{60}.\] Convert to a common denominator: \(\frac{10}{60} + \frac{18}{60} + \frac{13}{60} = \frac{41}{60}\).
03

Find P(5 < x ≤ 20)

\(P(5 < x \leq 20)\) is the sum of the probabilities for \(x = 10, 15, 20\): \[P(10) + P(15) + P(20) = \frac{7}{30} + \frac{1}{10} + \frac{3}{10}.\] Convert fractions to a common denominator: \(\frac{14}{60} + \frac{10}{60} + \frac{18}{60} = \frac{42}{60} = \frac{7}{10}\).
04

Find the Expected Value (E(X))

The expected value is calculated by multiplying each \(x\) value by its probability and summing the results: \[E(X) = 5\left(\frac{3}{20}\right) + 10\left(\frac{7}{30}\right) + 15\left(\frac{1}{10}\right) + 20\left(\frac{3}{10}\right) + 25\left(\frac{13}{60}\right).\] Simplifying each term and summing gives \(\frac{15}{20} + \frac{70}{30} + \frac{15}{10} + \frac{60}{10} + \frac{325}{60} = 15.08.\)
05

Find the Variance and Standard Deviation of X

Calculate for variance: \(V(X) = E(X^2) - \left(E(X)\right)^2\). First, find \(E(X^2)\) by summing \(x^2\) times the probability: \[E(X^2) = 5^2\left(\frac{3}{20}\right) + 10^2\left(\frac{7}{30}\right) + 15^2\left(\frac{1}{10}\right) + 20^2\left(\frac{3}{10}\right) + 25^2\left(\frac{13}{60}\right).\] This simplifies to 273.67. Then, \[V(X) = 273.67 - (15.08)^2 = 47.41.\] The standard deviation \(\sigma(X)\) is \(\sqrt{47.41} \approx 6.88.\)
06

Calculate E(Y) and V(Y)

The relationship between \(Y\) and \(X\) is \(Y = \frac{1}{5}X - 1\). Find \(E(Y)\) as \[E(Y) = \frac{1}{5}E(X) - 1 = \frac{1}{5}(15.08) - 1 = 2.016.\] For the variance, use \(V(Y) = \left(\frac{1}{5}\right)^2 V(X)\): \[V(Y) = \left(\frac{1}{5}\right)^2 \times 47.41 = 1.896.\] The standard deviation \(\sigma(Y)\) is \(\sqrt{1.896} \approx 1.376.\)

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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, often denoted as \( E(X) \), provides a measure of the center of a probability distribution, similar to the idea of a "weighted average". It tells us the average outcome we can expect if we repeated an experiment or certain process many times. To find the expected value of a discrete random variable \( X \), we use the formula: \[ E(X) = \sum_{i} x_i \cdot P(x_i), \]where \( x_i \) are the possible values the random variable can take, and \( P(x_i) \) are the associated probabilities. In the given exercise, calculating \( E(X) \) involves multiplying each value of \( x \) by its probability and summing all those products to get 15.08.
  • Firstly, identify all possible values of \( X \) and their probabilities.
  • Multiply each value of \( X \) by its corresponding probability.
  • Sum all these results to get the expected value.
Thus, the expected value gives us an insight into the average result or mean of the distribution.
Standard Deviation
Standard deviation, denoted as \( \sigma \), is a statistical measurement that illustrates the amount of variation or dispersion in a set of values. A low standard deviation means the values tend to be close to the mean, while a high standard deviation indicates the values are spread out over a wide range. It is derived from the square root of the variance, which we calculate once we know the expected value. To compute the standard deviation for \( X \):
  • First, calculate the variance \( V(X) \).
  • Find the square root of the variance using \( \sqrt{V(X)} \).
In our exercise, after computing the variance as 47.41, the standard deviation \( \sigma(X) \) is \( \sqrt{47.41} \approx 6.88 \).
Therefore, the standard deviation provides insight into how much variation or uncertainty there is within the probability distribution, and helps us understand the spread of possible outcomes.
Variance
Variance, represented as \( V(X) \), measures the degree of spread in a set of numbers. It essentially tells us how much the values in a data set deviate from the mean. To find the variance of a discrete random variable \( X \), follow these steps: \[ V(X) = E(X^2) - (E(X))^2 \]
  • Calculate \( E(X^2) \), which involves squaring each possible value of \( X \), multiplying by its probability, and summing these values.
  • Then compute \((E(X))^2 \), which is simply the square of the expected value.
  • The variance \( V(X) \) is the difference between \( E(X^2) \) and \((E(X))^2 \).
In the exercise, \( E(X^2) \) was calculated as 273.67. Subtracting the square of the expected value \((15.08)^2\), we determined the variance \( V(X) = 47.41 \).
Thus, variance is crucial for quantifying the degree of variation within a data set, serving as a foundational element for deeper statistical analysis.

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Most popular questions from this chapter

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