The top selling Red and Voss tire is rated 70,000 miles 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.
a. What is the probability that a tire wears out before 70,000 miles.
b. What is the probability that a tire last more than 100,000 miles?
X~N(82000,6400^2)
a) P(X<70000)=P((X-82000)/6400<(70000-82000¡ Z-table=0.0303964
P(X<70000)=0.0303964
b) P(X>100000)=P(Z>2.8125)=0.0024579
Please use the same name for your posts. Karen, Sandra, Joi.
The top-selling Amar tire is rated 70,000 KMs, 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 KMs and a standard deviation of 6,400 KMs.
What is the probability that a tire wears out before 70,000 KMs? What is the probability that a tire lasts more than 100,000 KMs?
Note: You may use Z-table for this
To find the probability in these situations, we will use the normal distribution and the Z-score formula. The Z-score measures the number of standard deviations a data point is from the mean.
a. To find the probability that a tire wears out before 70,000 miles, we need to calculate the Z-score first. The formula for the Z-score is:
Z = (X - μ) / σ
Where:
X = the value we want to find the probability of (70,000 miles in this case)
μ = the mean (82,000 miles)
σ = the standard deviation (6,400 miles)
Plugging the values into the formula:
Z = (70,000 - 82,000) / 6,400
Calculating the Z-score:
Z ≈ -1.875
Once we have the Z-score, we can use a Z-table or a calculator to find the corresponding probability.
Looking up the Z-score in the Z-table, we find that the probability associated with a Z-score of -1.875 is approximately 0.0301.
Therefore, the probability that a tire wears out before 70,000 miles is approximately 0.0301 or 3.01%.
b. To find the probability that a tire lasts more than 100,000 miles, we again need to calculate the Z-score using the same formula.
Z = (X - μ) / σ
Where:
X = the value we want to find the probability of (100,000 miles in this case)
μ = the mean (82,000 miles)
σ = the standard deviation (6,400 miles)
Plugging the values into the formula:
Z = (100,000 - 82,000) / 6,400
Calculating the Z-score:
Z ≈ 2.8125
Using the Z-table, we find that the probability associated with a Z-score of 2.8125 is approximately 0.9976.
Since we want the probability that the tire lasts more than 100,000 miles, we subtract this result from 1:
1 - 0.9976 ≈ 0.0024
Therefore, the probability that a tire lasts more than 100,000 miles is approximately 0.0024 or 0.24%.