Using the normal distribution, it is found that there is a 0.9192 = 91.92% probability that the total amount of product is less than 575 g.
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]
In this problem, the product is composed by flakes and raisins, and we have that:
[tex]\mu_F = 370, \sigma_F = 24, \mu_R = 170, \sigma_R = 7[/tex]
Hence, the distribution for the total amount of product has mean and standard deviation given by:
[tex]\mu = \mu_F + \mu_R = 370 + 170 = 540[/tex]
[tex]\sigma = \sqrt{\sigma_F^2 + \sigma_R^2} = \sqrt{24^2 + 7^2} = 25[/tex]
The probability that the total amount of product is less than 575 g is the p-value of Z when X = 575, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{575 - 540}{25}[/tex]
[tex]Z = 1.4[/tex]
[tex]Z = 1.4[/tex] has a p-value of 0.9192.
0.9192 = 91.92% probability that the total amount of product is less than 575 g.
A similar problem is given at https://brainly.com/question/22934264