### Understanding Normal Distribution in Monitor LifespanIn this example, we explore the lifespan of monitors manufactured by TSI Electronics, which are modeled with a normal distribution. The given parameters are:- **Variance**: 4,000,000 hours²- **Mean Lifespan**: 16,000 hoursOur objective is to find the probability that a randomly selected monitor will have a lifespan exceeding 14,800 hours. The solution requires answering a four-decimal place probability.#### Answer StrategyTo solve this, you typically use a standard normal distribution table or z-table. The steps are:1. **Calculate the standard deviation** by taking the square root of the variance: \[ \sigma = \sqrt{4,000,000} = 2,000 \text{ hours} \]2. **Find the z-score** for 14,800 hours: \[ z = \frac{X - \mu}{\sigma} = \frac{14,800 - 16,000}{2,000} = -0.6 \]3. **Use a z-table** to find the probability corresponding to this z-score.### Interactive Table ExplanationThe interface offers two normal distribution table options:- **Normal Table: -∞ to z**- **Normal Table: z to ∞****Instructions**:- Select the appropriate table.- Click the cell at the intersection of the row and column or use arrow keys for navigation. You can utilize space key selections for pinpoint accuracy in the table, enabling you to determine exact probability values.#### SubmissionOnce calculations are completed, submit your answer using the "Submit Answer" button provided.This educational tool helps students practice probability calculations using the normal distribution and enhances understanding through hands-on interaction with statistical tables.

It is given that the mean is 16,000 and the variance is 4,000,000.

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 4,000,000 and a mean life span of 16,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 14,800 hours. Round your answer to four decimal places.

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 4,000,000 and a mean life span of 16,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 14,800 hours. Round your answer to four decimal places.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Concept explainers

Continuous Probability Distributions

Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.

Normal Distribution

Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!

Question

### Understanding Normal Distribution in Monitor Lifespan

In this example, we explore the lifespan of monitors manufactured by TSI Electronics, which are modeled with a normal distribution. The given parameters are:

- **Variance**: 4,000,000 hours²
- **Mean Lifespan**: 16,000 hours

Our objective is to find the probability that a randomly selected monitor will have a lifespan exceeding 14,800 hours. The solution requires answering a four-decimal place probability.

#### Answer Strategy

To solve this, you typically use a standard normal distribution table or z-table. The steps are:

1. **Calculate the standard deviation** by taking the square root of the variance:
\[
\sigma = \sqrt{4,000,000} = 2,000 \text{ hours}
\]

2. **Find the z-score** for 14,800 hours:
\[
z = \frac{X - \mu}{\sigma} = \frac{14,800 - 16,000}{2,000} = -0.6
\]

3. **Use a z-table** to find the probability corresponding to this z-score.

### Interactive Table Explanation

The interface offers two normal distribution table options:

- **Normal Table: -∞ to z**
- **Normal Table: z to ∞**

**Instructions**:
- Select the appropriate table.
- Click the cell at the intersection of the row and column or use arrow keys for navigation.

You can utilize space key selections for pinpoint accuracy in the table, enabling you to determine exact probability values.

#### Submission

Once calculations are completed, submit your answer using the "Submit Answer" button provided.

This educational tool helps students practice probability calculations using the normal distribution and enhances understanding through hands-on interaction with statistical tables.

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