Chapter 5: Q 5.30 (page 214)

An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will be normally distributed with $μ = 4$ and $σ^{2} = 4$ , whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters $(6, 9)$ . A point is randomly chosen on the image and has a reading of $5$ . If the fraction of the image that is black is $α$ , for what value of $α$ would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region?

Short Answer

Expert verified

Therefore, the value of $α = 0.3827$ $α = 0.3827$ .

Step by step solution

Given information:

A reading taken from a randomly chosen point in the white section will be normally distributed with $μ = 4$ and $σ^{2} = 4$ , whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters $(6, 9)$

Explanation:

$P (g i v e n p o i n t b e l o n g t o b l a c k r e g i o n) = \frac{1}{2}$

Generally,

Equation 1:

$P (g i v e n p o i n t b e l o n g t o b l a c k r e g i o n) = \frac{α \times P (v a l u e = \frac{5}{b l a c k})}{α \times P (v a l u e = \frac{5}{b l a c k}) + (1 - α) \times P (v a l u e = \frac{5}{w h i t e})}$

Explanation:

Let's put them equal to each other:

Put equation 1 $\frac{α \times \frac{e^{\frac{- 1}{8}}}{2 \sqrt{(2 \times π)}}}{α \times \frac{e^{\frac{- 1}{8}}}{2 \sqrt{(2 \times π)}} + (1 - α) \frac{e^{\frac{- 1}{8}}}{3 \sqrt{(2 \times π)}}} = \frac{1}{2} \frac{\frac{α}{2}}{\frac{α}{2} + \frac{(1 - α) \times e^{\frac{- 5}{72}}}{3}} = \frac{1}{2} \frac{3 α}{3 α + 2 (1 - α) \times e^{\frac{- 5}{72}}} = \frac{1}{2} 6 α = 3 α + 2 (1 - α) \times e^{\frac{- 5}{72}} 3 α = 1.86 - 1.86 α α = 0.3827$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Given information:

Explanation:

Explanation:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Calculus

Theoretical and Mathematical Physics

Probability and Statistics

Statistics

Pure Maths

Study anywhere. Anytime. Across all devices.