Answer:

Step-by-step explanation:
We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find
.
<h3>First Case: items from 100 grams to the mean</h3>
For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".
A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.
Then

(1)
<h3>Second Case: items from the mean up to 190 grams</h3>
We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.
Then

(2)
<h3>Solving a system of equations for values of the mean and standard deviation</h3>
Having equations (1) and (2), we can form a system of two equations and two unknowns values:
(1)
(2)
Rearranging these two equations:
(1)
(2)
To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:
(1)
(2)
Summing both equations, we obtain the following equation:

Then

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):
(2)


