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

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: