Question

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is between 57,000 and 63,000 miles?

Answer #1

Given,

= 60000 , = 4000

Using central limit theorem,

P( < x) = P(Z < ( x - ) / ( / sqrt(n) ) )

So

P(57000 < < 63000 ) = P( < 63000 ) - P( < 57000 )

= P(Z < ( 63000 - 60000) / ( 4000 / sqrt(4) ) ) - P(Z < ( 57000 - 60000) / ( 4000 / sqrt(4) ) )

= P(Z < 1.5) - P(Z < -1.5)

= 0.9332 - 0.0668

= **0.8664**

The tread life of tires mounted on light-duty trucks follows the
normal probability distribution with a population mean of 60,000
miles and a population standard deviation of 4,000 miles. Suppose
we select a sample of 40 tires and use a simulator to determine the
tread life. What is the standard error of the mean?

The tread life of Road Stone tires has a normal distribution
with a mean of 35,000 miles and a standard deviation of 4,000
miles.
a. What proportion of these tires has a tread life of more than
38,000 miles?
b. What proportion of these tires has a tread life of less than
32,000 miles?
c. What proportion of these tires has a tread life of between
32,000 and 38,000 miles?

Suppose you work at a large tire distribution center. The tires’
average tread life has been 50,000 miles with a standard deviation
of 5,000 miles. At the end of the year, the company can reevaluate
their supply contract. There are four supply options for the next
contract: the current supplier or one of three competitors. The
current supplier produces tires with an average tread life of
50,000 miles with a standard deviation of 5,000 miles. Competitor A
claims to produce...

Part 3: The Normal Distribution
A tire manufacturer believes that the tread life of their
tires can be described by a normal distribution with a mean of
32,000 miles and a standard deviation of 2,500
miles.
Use StatCrunch. Copy and paste all StatCruch output
used.
What is the probability that one of these tires lasts
longer than 35,000 miles? (6 points)
What is the probability that one of these tires lasts
between 20,000 and 40,000 miles?
(6 points)
What is...

Suppose you work at a large tire distribution center. The tires’
average tread life has been 50,000 miles with a standard deviation
of 5,000 miles. At the end of the year, the company can reevaluate
their supply contract. There are four supply options for the next
contract: the current supplier or one of three competitors. The
current supplier produces tires with an average tread life of
50,000 miles with a standard deviation of 5,000 miles. Competitor A
claims to produce...

The tread life of a particular brand of tire is a random
variable best described by a normal distribution with a mean of
60,000 miles and a standard deviation of 2900 miles. What warranty
should the company use if they want 96% of the tires to outlast the
warranty?

The tread life of a particular brand of tire is a random
variable best described by a normal distribution with a mean of
60,000 miles and a standard deviation of 1400 miles. What is the
probability a certain tire of this brand will last between 56,850
miles and 57,300 miles?

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probability that a particular tire of this brand will last for
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Life of a tire follows normal distribution with a mean of 60,000
miles with standard deviation of 5000 miles. A tire of the same
kind is randomly selected. Find the probability that the tire will
last
(a) more than 52,000 miles
(b) less than 68,000 but more than 52,000 miles
(c) more than 68,000 miles.

The manufacturer of the X-15 steel-belted radial truck tire
claims that the mean mileage the tire can be driven before the
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the distribution is 4,000 miles. Crossett’s Truck Company bought 49
tires and found that the mean mileage for its trucks is 72,000
miles. is Crossett’s experience is different from that claimed by
the manufacturer at the 0.01 significance level

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