Math Problem Statement

Compute the probability that an observation 𝑦 drawn from a normal distribution with mean πœ‡=31.68 and standard deviation 𝜎=8.26 is in the interval (19.18,41.03), i.e., Prob(19.18<𝑦<41.03)

Solution

To compute the probability that an observation yy drawn from a normal distribution with mean ΞΌ=31.68\mu = 31.68 and standard deviation Οƒ=8.26\sigma = 8.26 lies in the interval (19.18,41.03)(19.18, 41.03), we'll use the properties of the normal distribution.

The probability P(19.18<y<41.03)P(19.18 < y < 41.03) can be calculated by finding the cumulative distribution function (CDF) values at the boundaries of the interval and then subtracting them.

Steps:

  1. Standardize the bounds: Convert the values 19.18 and 41.03 to their corresponding z-scores. z1=19.18βˆ’ΞΌΟƒ=19.18βˆ’31.688.26z_1 = \frac{19.18 - \mu}{\sigma} = \frac{19.18 - 31.68}{8.26} z2=41.03βˆ’ΞΌΟƒ=41.03βˆ’31.688.26z_2 = \frac{41.03 - \mu}{\sigma} = \frac{41.03 - 31.68}{8.26}

  2. Find the CDF for the z-scores: Use the standard normal distribution table (or a calculator) to find the probabilities corresponding to these z-scores.

  3. Compute the desired probability: P(19.18<y<41.03)=P(z1<Z<z2)=Ξ¦(z2)βˆ’Ξ¦(z1)P(19.18 < y < 41.03) = P(z_1 < Z < z_2) = \Phi(z_2) - \Phi(z_1) where Ξ¦(z)\Phi(z) is the CDF of the standard normal distribution.

Let's calculate this step by step.

Step 1: Standardize the bounds

z1=19.18βˆ’31.688.26β‰ˆβˆ’12.58.26β‰ˆβˆ’1.513z_1 = \frac{19.18 - 31.68}{8.26} \approx \frac{-12.5}{8.26} \approx -1.513 z2=41.03βˆ’31.688.26β‰ˆ9.358.26β‰ˆ1.132z_2 = \frac{41.03 - 31.68}{8.26} \approx \frac{9.35}{8.26} \approx 1.132

Step 2: Find the CDF values

Using a standard normal distribution table or calculator:

Ξ¦(βˆ’1.513)β‰ˆ0.0651\Phi(-1.513) \approx 0.0651 Ξ¦(1.132)β‰ˆ0.8716\Phi(1.132) \approx 0.8716

Step 3: Compute the probability

P(19.18<y<41.03)=Ξ¦(1.132)βˆ’Ξ¦(βˆ’1.513)β‰ˆ0.8716βˆ’0.0651=0.8065P(19.18 < y < 41.03) = \Phi(1.132) - \Phi(-1.513) \approx 0.8716 - 0.0651 = 0.8065

Final Answer

The probability that an observation yy lies in the interval (19.18,41.03)(19.18, 41.03) is approximately 0.8065.

Would you like more details on any part of this process or have any questions?


Here are five related questions you might explore:

  1. How do you calculate the probability of a z-score falling below a certain value?
  2. What is the relationship between the standard normal distribution and any normal distribution?
  3. How do you interpret the cumulative distribution function (CDF)?
  4. How would the probability change if the interval were narrower or wider?
  5. What are the applications of normal distribution in real-world scenarios?

Tip: Understanding z-scores is crucial in solving problems involving normal distributions as they allow us to use the standard normal distribution table for probability calculations.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Cumulative Distribution Function (CDF)
Z-Score

Formulas

Standard Normal Distribution Formula
Z-score Formula

Theorems

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Suitable Grade Level

Advanced High School