P(X >65)= P| H 65 – 50 10 = P(z>1.5) =1-P(z<1.5) [U sin g the excel function =1-0.9332 =NORM.DIST(1.5,0,1,TRUE) = 0.067

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
icon
Related questions
icon
Concept explainers
Topic Video
Question

I understand how you guy got 1.5 but after 1.5 how did you get 0.9332 and then the answer 0.067 can you break this problem down for me 

### Solving Probability Using the Standard Normal Distribution

To calculate the probability \( P(X > 65) \) for a normally distributed variable \( X \) with a mean (\(\mu\)) of 50 and a standard deviation (\(\sigma\)) of 10, follow these steps:

1. **Standardize the Variable**:
   Convert \( X \) to a standard normal variable \( z \) using the formula:
   \[
   P(X > 65) = P\left( \frac{X - \mu}{\sigma} > \frac{65 - 50}{10} \right)
   \]
   Substituting the values, we get:
   \[
   P(X > 65) = P(z > 1.5)
   \]

2. **Find the Complementary Probability**:
   To use standard normal distribution tables or functions, it's often easier to find the cumulative probability up to a point and subtract from 1:
   \[
   P(z > 1.5) = 1 - P(z < 1.5)
   \]
   
3. **Use Standard Normal Distribution Table or Function**:
   The cumulative probability \( P(z < 1.5) \) can be found using a standard normal distribution table or a computational tool such as Excel:
   \[
   1 - P(z < 1.5) \approx 1 - 0.9332
   \]
   This step involves using the Excel function:
   \[
   \text{Using the Excel function: } \text{=NORM.DIST}(1.5, 0, 1, \text{TRUE})
   \]
   Here, \( NORM.DIST(1.5, 0, 1, TRUE) \) yields approximately 0.9332.

4. **Calculate the Final Probability**:
   \[
   P(z > 1.5) = 1 - 0.9332 = 0.067
   \]

Therefore, the probability that \( X \) is greater than 65 is approximately \( 0.067 \). 

This approach utilizes the properties of the standard normal distribution to simplify the problem and allows for the efficient calculation using statistical software.
Transcribed Image Text:### Solving Probability Using the Standard Normal Distribution To calculate the probability \( P(X > 65) \) for a normally distributed variable \( X \) with a mean (\(\mu\)) of 50 and a standard deviation (\(\sigma\)) of 10, follow these steps: 1. **Standardize the Variable**: Convert \( X \) to a standard normal variable \( z \) using the formula: \[ P(X > 65) = P\left( \frac{X - \mu}{\sigma} > \frac{65 - 50}{10} \right) \] Substituting the values, we get: \[ P(X > 65) = P(z > 1.5) \] 2. **Find the Complementary Probability**: To use standard normal distribution tables or functions, it's often easier to find the cumulative probability up to a point and subtract from 1: \[ P(z > 1.5) = 1 - P(z < 1.5) \] 3. **Use Standard Normal Distribution Table or Function**: The cumulative probability \( P(z < 1.5) \) can be found using a standard normal distribution table or a computational tool such as Excel: \[ 1 - P(z < 1.5) \approx 1 - 0.9332 \] This step involves using the Excel function: \[ \text{Using the Excel function: } \text{=NORM.DIST}(1.5, 0, 1, \text{TRUE}) \] Here, \( NORM.DIST(1.5, 0, 1, TRUE) \) yields approximately 0.9332. 4. **Calculate the Final Probability**: \[ P(z > 1.5) = 1 - 0.9332 = 0.067 \] Therefore, the probability that \( X \) is greater than 65 is approximately \( 0.067 \). This approach utilizes the properties of the standard normal distribution to simplify the problem and allows for the efficient calculation using statistical software.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 1 images

Blurred answer
Knowledge Booster
Application of Algebra
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below.
Recommended textbooks for you
A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
A First Course in Probability
A First Course in Probability
Probability
ISBN:
9780321794772
Author:
Sheldon Ross
Publisher:
PEARSON