The lifetimes of the tread of a certain automobile tire are normally distributed with mean of 45,000 miles and a standard deviation of 3,000 miles.For each of the questions below make sure to show your work. Round any z-score calculated to two decimal places to find your result.What percent of tires have a tread life between 39,000 and 51,000 miles?The manufacturer wants to know what percent of times tread life is less than 37,500 miles.Find the tread life for the 90th percentile. Round to nearest integer (if needed)

Name: The lifetimes of the tread of a certain automobile tire are normally d
Uploaded: 2024-03-20T07:06:16.000Z
Channel: Bartleby
Description: The lifetimes of the tread of a certain automobile tire are normally distributed with mean of 45,000 miles and a standard deviation of 3,000 miles.For each of the questions below make sure to show your work. Round any z-score calculated to two decima

1. The percent of tires have a tread life between 39,000 and 51,000 miles is obtained below: From the given information, the value of mean is 45,000 miles and standard deviation is 3,000 miles. The required formula is given below: The required value is given below: In percentage = 0.9546 × 100% = 95.46% Thus, the percent of tires have a tread life between 39,000 and 51,000 miles is 95.46%.2. The percent of times tread life is less than 37,500 miles is obtained below: The required value is given below: In percentage = 0.0062 × 100% = 0.62% Thus, the percent of times tread life is less than 37,500 miles is 0.62%.3. The 90th percentile for the tread life is obtained below: The 90th percentile for the weight of seedless watermelons consider is P (Z≤z) = 0.90 From the standard normal table, locate the probability value of 0.90. Move to z until the first column is reached. Note the value as 1.2 Move upward until the top row is reached. Note the value as 0.08. The intersection of the row and column values gives the area to the two tail of z as 1.28. That is P (Z≤1.28) = 0.90.. The required value is, Thus, the 90th percentile for the tread life is 48,840.

Answered: The lifetimes of the tread of a certain…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

The lifetimes of the tread of a certain automobile tire are normally distributed with mean of 45,000 miles and a standard deviation of 3,000 miles.

For each of the questions below make sure to show your work. Round any z-score calculated to two decimal places to find your result.

What percent of tires have a tread life between 39,000 and 51,000 miles?
The manufacturer wants to know what percent of times tread life is less than 37,500 miles.
Find the tread life for the 90th percentile. Round to nearest integer (if needed)

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Centre, Spread, and Shape of a Distribution$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.

Similar questions