The amount of fill (weight of contents) put into a glass jar of spaghetti sauce is normally distributed with mean ì = 843 grams and standard deviation of ó = 9 grams

(ii) Find the standard error of the x distribution. (Give your answer correct to two decimal places.)
21.63 .

(e) Find the probability that a random sample of 20 jars has a mean weight between 836 and 855 grams. (Give your answer correct to four decimal places.)
0.1711

Your answers to both problems are wrong.

I am assuming your n is the 20 jars from the second question.

SEm = SD/√n = 9/√20 = 9/4.47 = ?

Z = (score-mean)/SEm

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.

To find the standard error of the x distribution, you'll need to use the formula:

Standard Error (SE) = Standard Deviation (σ) / Square Root of Sample Size (n)

In this case, the standard deviation (σ) is given as 9 grams, and the sample size (n) is 20 jars. Plugging these values into the formula, you can calculate the standard error as follows:

SE = 9 / √20

Using a calculator or by simplifying the square root, you'll find:

SE ≈ 9 / 4.472 ≈ 2.011

Rounding to two decimal places, the standard error is approximately 2.01 grams. Therefore, the answer to part (ii) is 2.01.

To find the probability that a random sample of 20 jars has a mean weight between 836 and 855 grams, you need to use the z-score and the standard normal distribution.

First, find the z-scores for the lower and upper bounds of the range:

Z1 = (836 - Mean (μ)) / Standard Error (SE) = (836 - 843) / 2.01 = -3.48
Z2 = (855 - Mean (μ)) / Standard Error (SE) = (855 - 843) / 2.01 = 5.97

Next, use a standard normal distribution table or calculator to find the area between these two z-scores (Z1 and Z2). Subtract the cumulative probability for Z1 from the cumulative probability for Z2 to find the probability of the mean weight falling within the given range.

P(836 < X < 855) = P(Z1 < Z < Z2)

Finding the probabilities individually:
P(Z < Z2) ≈ 1 (since Z2 is a large positive value, almost reaching the right tail of the distribution)
P(Z < Z1) ≈ 0 (since Z1 is a large negative value, almost reaching the left tail of the distribution)

Therefore, the probability of the mean weight falling between 836 and 855 grams is approximately:

P(836 < X < 855) ≈ P(Z1 < Z < Z2) = 1 - 0 = 1

Rounded to four decimal places, the probability is 1.0000. Therefore, the answer to part (e) is 0.1711.