The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean 43 and standard deviation 7. Find the percentile corresponding to p = 29% of the tire lifetimes.
To find the percentile corresponding to p = 29% of the tire lifetimes, we can use the standard normal distribution.
First, we need to find the z-score corresponding to the given percentile. The z-score represents the number of standard deviations a particular value is from the mean.
The formula for z-score is:
z = (x - μ) / σ
Where:
x = value we want to find the percentile for
μ = mean of the distribution
σ = standard deviation of the distribution
In this case, we have:
p = 29%
μ = 43 (mean)
σ = 7 (standard deviation)
To find the z-score, we rearrange the formula:
z = (x - μ) / σ
z = (x - 43) / 7
Next, we need to find the value of z that corresponds to a cumulative probability of 0.29. We can use a standard normal distribution table or a calculator to look up this value.
Looking up the z-value for a cumulative probability of 0.29, we find that z ≈ -0.5548.
Now, we can solve for x by rearranging the formula for z:
z = (x - 43) / 7
Plugging in the values we have:
-0.5548 = (x - 43) / 7
Solving for x:
-0.5548 * 7 = x - 43
-3.884 = x - 43
x = -3.884 + 43
x ≈ 39.116
Therefore, the value corresponding to the 29th percentile of tire lifetimes is approximately 39.116 thousand miles.
To find the percentile corresponding to a specific value in a normal distribution, you can use the z-score formula. The z-score measures the number of standard deviations a value is from the mean.
The formula for the z-score is: z = (x - μ) / σ
Where:
- z is the z-score
- x is the value you want to find the percentile for
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, we want to find the percentile corresponding to p = 29%, so we need to find the value of x.
First, let's calculate the z-score using the formula:
z = (x - μ) / σ
Substituting the given values:
z = (x - 43) / 7
Next, we want to find the value of x that corresponds to a z-score of 29%. To do this, we need to find the inverse of the standard normal distribution using a z-table or a statistical calculator.
For example, using a z-table, we can find that the z-score of 29% is approximately -0.5359.
Now, we can solve for x in the z-score formula:
-0.5359 = (x - 43) / 7
Cross multiplying, we get:
-3.7513 = x - 43
Adding 43 to both sides, we get:
x = 39.2487
So, the value of x corresponding to a percentile of 29% is approximately 39.2487.
Therefore, the 29th percentile of the tire lifetimes is approximately 39.2487 thousand miles.