Question

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

Answer 1

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 0.11702

From the complete question, we have:

$n = 41$ --- number of tires

$Mileage: 33095\ 34589\ 39411\ 42386\ 37886\ 33096\ 44185\ 38273\ 42387\ 36117$

$44373\ 39896\ 42758\ 34028\ 39768\ 44392\ 35826\ 44945\ 41756\ 41087$

$43716\ 33478\ 41430\ 39397\ 39517\ 38068\ 42216\ 43447\ 33372\ 42631$

$42215\ 44367\ 33186\ 41567\ 38534\ 33873\ 43484\ 39761\ 35531\ 40926\ 38348$

First, we calculate the mean

$\mu = \frac{\sum x}{n}$

This gives:

$\mu = \frac{33095+ 34589 +.............+40926 +38348}{41}$

$\mu = \frac{1619318}{41}$

$\mu = 39496$

Next, calculate the standard deviation

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

This gives:

$\sigma = \sqrt{\frac{(33095 - 39496)^2 + (34589 - 39496)^2 +.......+ (40926 - 39496)^2 + (38348- 39496)^2}{41-1}}$

$\sigma = \sqrt{\frac{572531448}{40}}$

$\sigma = \sqrt{14313286.2}$

$\sigma = 3783$

The probability a tire will fail before 35000 is represented as:

$P(x < 35000)$

Calculate the z score

$z = \frac{x - \mu}{\sigma}$

This gives

$z = \frac{35000 - 39496}{3783}$

$z = \frac{-4496}{3783}$

$z = -1.19$

So, we have:

$P(x < 35000) = P(z < -1.19)$

From z table of values:

$P(z < -1.19) = 0.11702$

Hence:

$P(x < 35000) = 0.11702$

So, the probability that the selected tire fails before 35000-mile warranty is 0.11702

Read more about normal distribution probabilities at:

https://brainly.com/question/16464000