A tire company designed a new tire tread that was considered safe in all weather

Question

A tire company designed a new tire tread that was considered safe in all weather conditions. The tires were tested on vehicles that had to stop completely after reaching a speed of 50 miles per hour. All vehicles weighed roughly the same. The mean stopping distance was 110 feet with a standard deviation of 5.8 feet. A review of data indicated that the stopping distances were normally distributed.

1.) What stopping distance corresponds to the 75th percentile?

2.) If 5 vehicles are selected at random, what is the probability that at least two of them will exceed the distance that comprises the 75th percentile?

3.) What is the probability that a randomly selected sample of five cars will have a mean stopping distance of less than 105 feet?

If possible, please post at least the equation or calculator function used to solve these problems. Mostly the calculator function is needed and would be the most helpful! Thank you!!

Explanation / Answer

1) mean = 110

standard deviation = 5.8

the zscore for 75th percentile = 0.675

x = 110+0.675*5.8

= 113.915

b) p(x>113.915) = z = (113.915-110)/5.8 = 0.675

hence p = 0.75

p(x>=2) = 1 - [p(0)+p(1)]

p(0) = 0.25^5 = 0.0009

p(1) = 5C1*(0.75)^1*(0.25)^4 = 0.0146

HENCE P(X>=2) = 1 - 0.0155 = 0.9845

C) P(X>105) = Z = (105-110)/2.8 = -1.78

Z = -1.78

THEREFORE P = 0.0375

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.