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 %
Dear Hay, please use the panel for submitting new questions.
Hay
13.12.20, 14:30
Question 1 (a). A casino gambler keeps on inserting fifty Ghana pesewa
coins into a vending machine with the hope of winning ten Ghana cedis
after each try. After five tries, he has not won anything.
Nevertheless, he decides to continue playing the game. (i). Indicate
whether the decision taken is right or wrong, giving reasons for your
answer. (ii). Base on your decision taken in (i) above, explain the
probability concept that the gambler employed in playing the game.
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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Dear Hay, please use the panel for submitting new questions.
Question 1 (a). A casino gambler keeps on inserting fifty Ghana pesewa coins into a vending machine with the hope of winning ten Ghana cedis after each try. After five tries, he has not won anything. Nevertheless, he decides to continue playing the game. (i). Indicate whether the decision taken is right or wrong, giving reasons for your answer. (ii). Base on your decision taken in (i) above, explain the probability concept that the gambler employed in playing the game.
Yes, you're absolutely right. Thanks for correcting us!
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?
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.
how do you know P (x>50,000) = .62%? I am very confused when it comes to the P (X) part
Leave a comment