Mathematics, 26.04.2020 05:42, mcdowelle360

PLEASE HELP!!
1) Machines that fill bags with powdered sugar, also called confectioner’s sugar, are
supposed to dispense exactly 32 ounces of powdered sugar into each bag. Let X = the weight in ounces of
the powdered sugar dispensed into a randomly selected bag. Suppose that X follows a normal distribution
with mean 32 ounces and standard deviation 0.6 ounces. Find and interpret P(X  31).

2) Refer to the previous problem. What percent of all bags contain 31 to 33 ounces of
powdered sugar?

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answered: maderaada

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 $P(X \le 31)$ is approximately the same as finding $P(Z \le -1.67)$

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, $P(Z \le -1.67) \approx 0.0475$ and that $P(X \le 31) \approx 0.0475$

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

$P(Z \le 1.67) \approx 0.9525$

therefore,

$P(X \le 33) \approx 0.9525$

and,

$P(A \le X \le B) = P(X \le B) - P(X \le A)\\\\P(31 \le X \le 33) = P(X \le 33) - P(X \le 31)\\\\P(31 \le X \le 33) \approx 0.9525 - 0.0475\\\\P(31 \le X \le 33) \approx 0.905$

Therefore, the percent of all bags with weights between 31 and 33 ounces is roughly 90.5%

PLEASE HELP!! 1) Machines that fill bags with powdered sugar, also called confectioner’s sugar, are

answered: Guest

If you make it a little more clear i can answer it

answered: Guest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$