Question

4. Interpreting area under the standard normal curve Suppose Dunlop Tire manufactures tires having the property that the mileage the tire lasts approximately follows a normal distribution with mean u=70,000 miles and a standard deviation o=4400 miles. a. What percent of the tires will last at least 75,000? b. Suppose Dunlop gives a warranty for tires to last 60,000 miles. What percent of the tires will last 60,000miles or less? c. What is the probability that a randomly selected Dunlop tire lasts between 65,000 and 80,000 miles? (Hint: find the z-scores, then use the standard normal table to find the area under the curve. Pay attention to whether the question is asking for the area to the left or to the right or in between)

Answer 1

(a) Solution: 0.1279 Working: Find P(X2 75000). Normal Distribution, u =70000, 0 =4400 In the enclosed graph, the shaded region P(X Thus, P(XX 75000)=0.1279022=0.1279 75000) with u =70000 and 0 =4400 (6) Solution: 0.0115 Working: Find P(X360000). Normal Distribution, u =70000, 0 =4400 In the enclosed graph, the shaded region P(X360000) with u =70000 and 0 =4400 Thus, P(X360000)=0.01152131 -0.0115 (c) Solution: 0.8606 Working: Find P(65000SX 80000). Normal Distribution, u =70000, 0 =4400 In the enclosed graph, the shaded region P(60000 SX 85000) with u =70000 and 0 =4400. Thus, P(60000 SX 85000) =0.86057649 -0.8606

f(x) GRAPH-1 GRAPH-2 0.00008 0.00008 0.00006 0.00006 0.00004 0.000041 0.00002 0.00002 0 60000 65000 70000 75000 80000 60000 65000 70000 75000 80000 f(x) GRAPH-3 0.000081 0.00006 0.000041 0.000021 ot 60000 65000 70000 75000 80000 X