The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean =41 and standard deviation =6
(a) What is the probability that a randomly chosen tire has a lifetime greater than 50 thousand miles?

6.68% probability that a randomly chosen tire has a lifetime greater than 50 thousand miles

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

(a) What is the probability that a randomly chosen tire has a lifetime greater than 50 thousand miles?

This is 1 subtracted by the pvalue of Z when X = 50. So

has a pvalue of 0.9332

1 - 0.9332 = 0.0668

6.68% probability that a randomly chosen tire has a lifetime greater than 50 thousand miles

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

won't that be plagrisim thought?

step-by-step explanation: