A maker of tires for cars has a multitude of data to support the fact that the lifetime of their tires follows a normal distribution with a mean of 42,100 miles and a standard deviation of 2,510 miles.
Find the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles. Be certain that you round your z-values to two decimal places. Round your answer to 4 decimal places.
This is what I got:
0.3315
Am I correct?
I got .1601
by using
http://davidmlane.com/hyperstat/z_table.html
just enter the values from your problem.
To find the probability that a randomly selected tire will have a lifetime between 44,500 miles and 48,000 miles, we first need to standardize these values using the z-score formula.
The z-score formula is calculated as (x - mean) / standard deviation, where x is the value we want to standardize, mean is the mean of the distribution, and standard deviation is the standard deviation of the distribution.
For 44,500 miles:
z1 = (44,500 - 42,100) / 2,510 = 0.956
For 48,000 miles:
z2 = (48,000 - 42,100) / 2,510 = 2.35
Next, we need to find the area under the normal distribution curve between these two z-scores. We can use a standard normal distribution table or a statistical calculator to find these probabilities.
Using a standard normal distribution table, we can find the corresponding probabilities for z1 and z2.
The probability corresponding to z1 is approximately 0.8315, and the probability corresponding to z2 is approximately 0.9907.
To find the probability between these two z-scores, we subtract the probability corresponding to z1 from the probability corresponding to z2:
P(44,500 < x < 48,000) = P(z1 < z < z2) = P(z < z2) - P(z < z1)
P(44,500 < x < 48,000) ≈ 0.9907 - 0.8315 = 0.1592
Therefore, the probability that a randomly selected tire will have a lifetime between 44,500 miles and 48,000 miles is approximately 0.1592.
Your answer of 0.3315 is incorrect; the correct answer is approximately 0.1592, rounded to 4 decimal places.