Dunlop makes tires that the mileage the tire lasts follows a normal distribution with the mean 65,000 miles and standard deviation of 4600 miles.

Question

What is the probability that the randomly selected tire will last at least 70,000 miles?

Suppose Dunlop warrants the tires for 60,000 miles, what proportion of the tires will last 55,000 miles or less?

Suppose a quality control engineer obtains a simple random sample of n=10 tires and calculates the mean mileage for the 10 tires. Describe the distribution that x(bar/mean) the sample mean belongs to. Shape, center, spread.

Answer 1

Compute the z-score z = (x - mean) / (standard deviation) 

 

Z(70000) = (70000 - 65000) / 4600 = 1.09  

 

Then look up the z-score in a Z table or use a statistical calculator 

 

P(Z > 1.09) = 1 - P(Z < 1.09) = 1 - 0.8621 = 0.1379

 

Z(55000) = (55000 - 65000) / 4600 = -2.17

 

P(Z < -2.17) = 0.0150 

 

Since the population distribution is normal the sample distribution will also be normal with mean 65000 and standard deviation 4600/√10 = 1454.6