Answer to Question #840 in Statistics and Probability for jasmine

Question #840
A new Firestone tire is guaranteed to last for 40,000 miles. The actual mean life of the tires is 47,000 miles with a standard deviation of 4,000 miles. a) What percent of the tires will last for at least 40,000 miles? b) What percent of the tires will not last for at least 40,000 miles? c) What is the probability that a tire will last for more than 50,000 miles? d) The Firestone Company wants to advertise how long some of their tires last. They decide to state how long the top 3% of their tires will last. How many miles will the top 3% of their tires last?
1
Expert's answer
2010-10-24T04:11:57-0400
a) Let's find z - score:
z = (x-mu)/sigma = (40,000-47,000)/4,000 = - 1.75
p-value for z = 1.75 is 0.0401, for estimating the percent of the tires which last for at least 40,000 miles we have to consider right side of the distribution:
P(x > 40,000) = (1-P(x < 40,000))*100% = 95.99 %


b) the percent of the tires which are not last for at least 40,000 miles would be equal to P (x < 40,000) = 0.0401*100% = 4.01 %;
c) for more than 50,000 miles
z = (50,000-40,000)/4,000 = 2.5
the probability would be
P(x>50,000) = 00.62 %

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Comments

Assignment Expert
14.08.12, 15:58

Yes, you're absolutely right. Thanks for correcting us!

Melody
13.08.12, 20:32

Why, in answer c, is the z-score formula z=(50000-40000)/4000? Because the mean is 47000, should it be z=(50000-47000)/4000?

Assignment Expert
05.04.12, 14:36

Normal distribution with given mean and variance is considered for solving the problem. For example, take a look https://en.wikipedia.org/wiki/Standard_normal_table
The standard normal distribution table provides the probability that random variable x is less than or equal to z. It does this for positive values of z only. So from given table we can find values of& P(x).
So values of p(x) are taken from the table of such kind.

Julian
04.04.12, 04:44

how do you know P (x>50,000) = .62%? I am very confused when it comes to the P (X) part

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