The tread lives of the Super Titan radial tires under normal driving conditions are normally distributed with a mean of 34000 mi and a standard deviation of 1700 mi.
(a) What is the probability that a tire selected at random will have a tread life of more than 35071 mi? Round your standard normal variable to two decimal places before using the table of values. (Give your answer to four decimal places.)
(b) If four new tires are installed in a car and they experience even wear, determine the probability that all four tires still have useful tread lives after 35071 mi of driving. Round your standard normal variable to two decimal places before using the table of values. (Give your answer to four decimal places.)
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To solve these probability problems, we need to use the concept of standard normal distribution, also known as the z-distribution. The z-score for a given value x represents how many standard deviations x is away from the mean.
(a) To find the probability that a tire will have a tread life of more than 35071 mi, we need to calculate the area to the right of this value on the standard normal distribution curve.
First, we need to calculate the z-score for 35071 mi:
z = (x - mean) / standard deviation
z = (35071 - 34000) / 1700
z = 1.14 (rounded to two decimal places)
Next, we can use a standard normal distribution table or calculator to find the corresponding probability. The table provides the proportion of the curve's area to the right of the z-score.
Looking up the z-score of 1.14 in the table, we find that the area to the right is approximately 0.1271.
Therefore, the probability that a tire selected at random will have a tread life of more than 35071 mi is 0.1271 (rounded to four decimal places).
(b) To determine the probability that all four tires still have useful tread lives after 35071 mi of driving, we need to calculate the probability for a single tire and then raise it to the power of four, since the events are independent.
Using the same formula as in part (a), we find the z-score for 35071 mi:
z = (35071 - 34000) / 1700
z = 1.14 (rounded to two decimal places)
Using the standard normal distribution table, we find that the area to the right of the z-score of 1.14 is approximately 0.1271.
Since the events are independent, the probability that all four tires still have useful tread lives after 35071 mi is calculated by raising the probability for a single tire to the power of four:
P(all four tires have useful tread lives) = 0.1271^4
P(all four tires have useful tread lives) ≈ 0.0021 (rounded to four decimal places).
Therefore, the probability that all four tires still have useful tread lives after 35071 mi of driving is approximately 0.0021 (rounded to four decimal places).