In: Statistics and Probability
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 47,225 miles, with a standard deviation of 3178 miles. What is the probability that the sample mean would be greater than 47,050 miles in a sample of 208 tires if the manager is correct? Round your answer to four decimal places.
Solution :
Given that ,
mean = = 47225
standard deviation = = 3178
= / n = 3178 / 208 = 220.3547
P( > 47050) = 1 - P( < 47050)
= 1 - P[( - ) / < (47050 - 47225) / 220.3547]
= 1 - P(z < -0.7942)
= 1 - 0.2135
= 0.7865
Probability = 0.7865