Using the normal distribution, it is found that there is a 0.6406 = 64.06% probability that the company will have to refund money to a customer.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Researching the problem on the internet, it is found that the mean and the standard deviation are, respectively, given by [tex]\mu = 13[/tex] and [tex]\sigma = 2.75[/tex].
The probability it lasts less than 14 years is the p-value of Z when X = 14, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{14 - 13}{2.75}[/tex]
[tex]Z = 0.36[/tex]
[tex]Z = 0.36[/tex] has a p-value of 0.6406.
0.6406 = 64.06% probability that the company will have to refund money to a customer.
More can be learned about the normal distribution at https://brainly.com/question/24663213