The tread life of a particular tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 2000 miles. What is the probability that a randomly selected tire of this brand will last longer than 58,000 miles?
A.0.7266, B.0.8413, C.0.1587, D.0.2266
A.0.7266
To find the probability that a randomly selected tire of this brand will last longer than 58,000 miles, we need to calculate the area under the normal distribution curve to the right of 58,000 miles.
First, we need to standardize the value of 58,000 miles by using the z-score formula:
z = (x - μ) / σ
where:
x = value (58,000 miles)
μ = mean (60,000 miles)
σ = standard deviation (2,000 miles)
Substituting the values:
z = (58,000 - 60,000) / 2,000
z = -1,000 / 2,000
z = -0.5
Now, we can use a standard normal distribution table or a calculator to find the area to the right of z = -0.5. The value will represent the probability that a randomly selected tire will last longer than 58,000 miles.
Looking up the z-score of -0.5 in the table, we find that the area to the right is 0.6915.
However, since we want the probability to the right of -0.5, we need to subtract this value from 1 to get the desired probability:
Probability = 1 - 0.6915
Probability = 0.3085
Therefore, the probability that a randomly selected tire of this brand will last longer than 58,000 miles is approximately 0.3085, which rounds to 0.309. Therefore, the answer is not one of the given options (A, B, C, D).
To find the probability, we need to standardize the value of 58,000 miles using the given information about the mean and standard deviation.
The standardized value, also known as the z-score, can be found using the formula:
z = (x - mean) / standard deviation
In this case, x (the given value) is 58,000 miles, the mean is 60,000 miles, and the standard deviation is 2000 miles.
Substituting these values:
z = (58,000 - 60,000) / 2000
z = -2000 / 2000
z = -1
Now, we need to find the probability that a randomly selected tire will last longer than 58,000 miles. Since we are interested in the area to the right of the z-score of -1, we can look up this probability in the standard normal distribution table (also known as the Z-table).
Looking up the z-score of -1 in the table, we find that the corresponding probability is 0.1587.
Therefore, the answer is C. 0.1587.