Answer:
Explanation:
We know that the life expectancy of the tires is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. Let X be the life expectancy of the tire. Then X ~ N(40,000, 5,000^2).
To find the percentage of tires that will have a life of 34,000 to 46,000 miles, we need to calculate the z-scores for both values using the formula:
z = (X - μ) / σ
where X is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation.
For X = 34,000:
z = (34,000 - 40,000) / 5,000 = -1.2
For X = 46,000:
z = (46,000 - 40,000) / 5,000 = 1.2
Now, we need to find the area under the standard normal distribution curve between these two z-scores. We can use a table of standard normal probabilities or a calculator to find this area.
Using a standard normal table or a calculator, we find that the area to the left of z = -1.2 is 0.1151, and the area to the left of z = 1.2 is 0.8849. Therefore, the area between these two z-scores is:
0.8849 - 0.1151 = 0.7698
So, approximately 76.98% of the tires will have a life of 34,000 to 46,000 miles.