10th Edition

ISBN: 9780134672090

Author: Demana

Publisher: PEARSON

Expert Solution & Answer

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.

Chapter 10.4, Problem 38E

Solution

a.

To find: The percent of tires wear out before they reach 40,000 mi.

The percent of tires wear out before they reach 40,000 mi is 85.5%.

Given information:

Mean μ=37200

Standard deviation σ=2650

Calculation:

Assume the life of tire follows a normal distribution.

Let X be the life of tire.

Compute the value of z for X=40000 .

z=X−μσ=40000−372002650=1.0566

The probability is:

PX<40000=Px<1.0566=0.8546

Thus, the percent of tires wear out before they reach 40,000 mi is 85.5%.

b.

To find: The probability a tire will wear out between 32,000 and 35,000 mi.

The probability a tire will wear out between 32,000 and 35,000 mi is 17.8%.

Calculation:

Assume the life of tire follows a normal distribution.

Let X be the life of tire.

Compute the value of z for X=32000 .

z=X−μσ=32000−372002650=−1.9622

Compute the value of z for X=35000 .

z=X−μσ=35000−372002650=−0.83

The probability is:

P32000<X<35000=P(−1.9622<z<−0.83)=P−0.83−P(−1.9622)=0.2032694−0.0248696 (By calculator)≈0.178

Thus, the probability a tire will wear out between 32,000 and 35,000 mi is 17.8%.

c.

To find: The number of miles can the company guarantee these tires, if the company wants at least 90% of them to last that long.

The required life of tires is about 33,800 miles.

Calculation:

Using the calculator’s inverse Normal function, we find the 90th percentile at z= 1.2815515 .

This represents the life of tire that is 1.2815515 standard deviation below the mean.

The life of tires could be calculated as:

x=μ+z×σ=37200+1.2815515×2650≈40596

Hence, the required life of tires is about 40596 miles.

Chapter 10.4, Problem 37E

Chapter 10.4, Problem 39E

Chapter 10 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

Ch. 10.1 - Prob. 1QR Ch. 10.1 - Prob. 2QR Ch. 10.1 - Prob. 3QR Ch. 10.1 - Prob. 4QR Ch. 10.1 - Prob. 5QR Ch. 10.1 - Prob. 6QR Ch. 10.1 - Prob. 7QR Ch. 10.1 - Prob. 8QR Ch. 10.1 - Prob. 9QR Ch. 10.1 - Prob. 10QR

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.1 - Prob. 33E Ch. 10.1 - Prob. 34E Ch. 10.1 - Prob. 35E Ch. 10.1 - Prob. 36E Ch. 10.1 - Prob. 37E Ch. 10.1 - Prob. 38E Ch. 10.1 - Prob. 39E Ch. 10.1 - Prob. 40E Ch. 10.1 - Prob. 41E Ch. 10.1 - Prob. 42E Ch. 10.1 - Prob. 43E Ch. 10.1 - Prob. 44E Ch. 10.1 - Prob. 45E Ch. 10.1 - Prob. 46E Ch. 10.1 - Prob. 47E Ch. 10.1 - Prob. 48E Ch. 10.1 - 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