The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 39$ and standard deviation $\sigma = 6$ . What proportion of tires have lifetimes greater than 48 Thousand miles?
A) 0.4666
B) 0.0668
C) 0.5334
D) 0.9332

Question

Quizplus · Accepted Answer

We need to find $P(X > 48)$, where $X$ is a normally distributed random variable with $\mu = 39$ and $\sigma = 6$.

Using z-score, we can standardize the value of 48:
$$z = \frac{48 - 39}{6} = 1.5$$

Now we look up the probability of z being less than 1.5 in the standard normal distribution table which gives us 0.9332.

Since we want the probability that $X$ is greater than 48, we subtract this value from 1:
$$P(X > 48) = 1 - 0.9332 = 0.0668$$

Therefore, the answer is B.

The Lifetime of a Certain Type of Automobile Tire (In μ=39\mu = 39μ=39

The Lifetime of a Certain Type of Automobile Tire (In $\mu = 39$