Problem 1
mu = 32 is the population mean, sigma = 0.6 is the population standard deviation
Convert the raw score x = 31 into its z score
z = (x-mu)/sigma
z = (31-32)/0.6
z = -1.67 approximately
Finding [tex]P(X \le 31)[/tex] is approximately the same as finding [tex]P(Z \le -1.67)[/tex]
To find this value, we use a calculator or a table. I'll use a table. Specifically it is called a Z table. They are found in the back of your stats textbook. If you don't have a textbook with you, then they can be found freely online by searching out "z table".
Check out the attached image below. In that image I have marked the row that starts with -1.6 and the column that has 0.07 at the top. These two values combine to -1.67, with the value 0.0475 at the intersection of this row and column.
This means, [tex]P(Z \le -1.67) \approx 0.0475[/tex] and that [tex]P(X \le 31) \approx 0.0475[/tex]
Interpretation: The probability of getting a bag of weight 31 ounces or less is approximately 0.0475 (which is a 4.75% chance roughly).
Another interpretation: Roughly 4.75% of all the bags have a weight of 31 ounces or less.
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Problem 2
Earlier we found that x = 31 has a z score of roughly z = -1.67
Convert x = 33 to its corresponding z score and you should find that it is roughly z = 1.67
Use the Z table to find that
[tex]P(Z \le 1.67) \approx 0.9525[/tex]
therefore,
[tex]P(X \le 33) \approx 0.9525[/tex]
and,
[tex]P(A \le X \le B) = P(X \le B) - P(X \le A)\\\\P(31 \le X \le 33) = P(X \le 33) - P(X \le 31)\\\\P(31 \le X \le 33) \approx 0.9525 - 0.0475\\\\P(31 \le X \le 33) \approx 0.905[/tex]
Therefore, the percent of all bags with weights between 31 and 33 ounces is roughly 90.5%
