Question

A popcorn company builds a machine to fill 1 kg bags of popcorn. They test the first hundred bags filled and find that the bags have an average weight of 1,040 grams with a standard deviation of 25 grams. 1.) Fill out the normal distribution curve for this situation. 2.) What percentage of people would receive a bag that had a weight greater than 1115 grams?

Answer 1

To fill out the normal distribution curve, use the average weight and standard deviation. To find the percentage of people who would receive a bag with a weight greater than a certain amount, calculate the Z-score and use a Z-table.

Step-by-step explanation:

To fill out the normal distribution curve for this situation, we can use the given information. The bags have an average weight of 1,040 grams with a standard deviation of 25 grams. This means that the mean of the distribution is 1,040 and the standard deviation is 25. We can plot the normal distribution curve using these values.

To find the percentage of people who would receive a bag that had a weight greater than 1,115 grams, we need to find the area under the normal distribution curve to the right of 1,115. We can calculate this using a Z-score calculator or a Z-table. The Z-score for 1,115 is (1,115 - 1,040) / 25 = 3. From the Z-table, we can find that the area to the right of Z-score 3 is approximately 0.0013. This means that approximately 0.13% of people would receive a bag that weighs more than 1,115 grams.

Answer 2

Step-by-step explanation:

Given that a popcorn company builds a machine to fill 1 kg bags of popcorn. They test the first hundred bags filled and find that the bags have an average weight of 1,040 grams with a standard deviation of 25 grams.

i.e. Sample mean = 1040 and

Sample std dev s = 25 gm

Sample size n = 100

Hence by central limit theorem we have the sample mean follows a normal distribution with mean =1040 and std dev = s = 25 gm

`\bar X = N(1040,25)`

Normal curve would be with mean 1040 and std deviatin 25

b) P(X>1115)

= 1-0.9987

=0.0013

i.e. 0.13% would receive a bag that had a weight greater than 1115 grams