Please help I was up till 2am trying to figure this out last night, Thanks
The top selling Red and Voss is rated 70,000 miles, which means nothing. In fact, the distance the tires can run until they wear out is a normally distributed random variable with a mean of 82,000 miles and a standard deviation of 6,400 miles.
a) What is the probability that the tire wears out before 70,000 miles?
b) What is the probability that the tire last more than 100,000miles?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to your Z scores.
Thank you PSyDAG
To solve this problem, we can use the concept of standard normal distribution and the z-score.
a) To find the probability that the tire wears out before 70,000 miles, we need to find the area under the standard normal curve to the left of the z-score corresponding to 70,000 miles.
Step 1: Calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = 70,000 miles
μ = mean = 82,000 miles
σ = standard deviation = 6,400 miles
Plugging in the values:
z = (70,000 - 82,000) / 6,400
z = -12,000 / 6,400
z ≈ -1.875
Step 2: Use a z-table or a calculator to find the area under the standard normal curve to the left of -1.875. This area represents the probability.
Using a z-table, we can find the probability as approximately 0.0301 (or 3.01%).
Therefore, the probability that the tire wears out before 70,000 miles is approximately 0.0301 or 3.01%.
b) To find the probability that the tire lasts more than 100,000 miles, we need to find the area under the standard normal curve to the right of the z-score corresponding to 100,000 miles.
Step 1: Calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = 100,000 miles
μ = mean = 82,000 miles
σ = standard deviation = 6,400 miles
Plugging in the values:
z = (100,000 - 82,000) / 6,400
z = 18,000 / 6,400
z ≈ 2.8125
Step 2: Use a z-table or a calculator to find the area under the standard normal curve to the right of 2.8125. This area represents the probability.
Using a z-table, we can find the probability as approximately 0.0025 (or 0.25%).
Therefore, the probability that the tire lasts more than 100,000 miles is approximately 0.0025 or 0.25%.