Q:

What is the probability that a randomly selected tire will fail before the 35,000 mile warranty mileage stated?

Accepted Solution

A:
Probabilities are used to determine how likely, or often an event is, to happen. The probability that the selected tire fails before 35000-mile warranty is 0.11702From the complete question, we have:[tex]n = 41[/tex] --- number of tires[tex]Mileage: 33095\ 34589\ 39411\ 42386\ 37886\ 33096\ 44185\ 38273\ 42387\ 36117[/tex][tex]44373\ 39896\ 42758\ 34028\ 39768\ 44392\ 35826\ 44945\ 41756\ 41087[/tex][tex]43716\ 33478\ 41430\ 39397\ 39517\ 38068\ 42216\ 43447\ 33372\ 42631[/tex][tex]42215\ 44367\ 33186\ 41567\ 38534\ 33873\ 43484\ 39761\ 35531\ 40926\ 38348[/tex]First, we calculate the mean[tex]\mu = \frac{\sum x}{n}[/tex]This gives:[tex]\mu = \frac{33095+ 34589 +.............+40926 +38348}{41}[/tex][tex]\mu = \frac{1619318}{41}[/tex][tex]\mu = 39496[/tex]Next, calculate the standard deviation[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]This gives:[tex]\sigma = \sqrt{\frac{(33095 - 39496)^2 + (34589 - 39496)^2 +.......+ (40926 - 39496)^2 + (38348- 39496)^2}{41-1}}[/tex][tex]\sigma = \sqrt{\frac{572531448}{40}}[/tex][tex]\sigma = \sqrt{14313286.2}[/tex][tex]\sigma = 3783[/tex]The probability a tire will fail before 35000 is represented as:[tex]P(x < 35000)[/tex]Calculate the z score[tex]z = \frac{x - \mu}{\sigma}[/tex]This gives[tex]z = \frac{35000 - 39496}{3783}[/tex][tex]z = \frac{-4496}{3783}[/tex][tex]z = -1.19[/tex]So, we have:[tex]P(x < 35000) = P(z < -1.19)[/tex]From z table of values:[tex]P(z < -1.19) = 0.11702[/tex]Hence:[tex]P(x < 35000) = 0.11702[/tex]So, the probability that the selected tire fails before 35000-mile warranty is 0.11702Read more about normal distribution probabilities at: