Oxford Cereal

Statistics

Problem 1: Oxford Cereals

Q1

The sample mean

First sample (Oxford O)

= (360.4 + 361.8 + 362.3 + 364.2 + 371.4) / 5

= 1820.1 / 5

= 364.02 grams

Second sample (Alpine Granola Frosted Flakes)

= (366.1 + 367.2 + 365.6 + 367.8 + 373.5) / 5

= 1840.2 / 5

= 368.04 grams

The percentage of the sample mean

Oxford O

ZCalculated = (mean of the sample – population mean) / standard error

= (364.02 – 368)

15 / √5

= -3.98 / 6.708

Z = -0.5933

The area corresponding to Z = -0.5933 is 0.2776. This implies that 27.76% of all the samples have a weight that is less than 364.02 grams.

Alpine Granola Frosted Flakes

ZCalculated = (368.04 – 368)

15 / √5

= 0.04 / 6.708

Z = 0.00596

The area corresponding to Z = 0.01 is 0.496. This implies that 49.6% of the sample has a weight that is less than 368.04 grams.

The percentage of individual boxes

Oxford O

ZCalculated = (Sample mean – population mean) / standard deviation

= (364.02 – 368)

15

= -3.98 / 15

Z = -0.2653

The area corresponding to Z = -0.27 is 0.3936. This implies that 39.36% of the individual boxes of cereal have a weight that is less than 364.02 grams.

Alpine Granola Frosted Flakes

ZCalculated = (368.04 – 368)

15

= 0.04 / 15

Z = 0.00266

The area corresponding to Z = 0.00 is 0.5000. This implies that 50% of the individual boxes of cereal have a weight that is less than 368.04 grams.

Q2

Based on the calculations, it can be observed that a larger percentage of individual boxes have weights below population mean than the sample means. This can be attributed to variations in weights of individual boxes. Therefore, the probability that an individual box is far from the population mean is greater than the probability that the sample mean is far from the population mean.

Q3

The request is not reasonable since there is sufficient evidence, based on the statistical calculations, that a large percentage of the weight of the individual boxes is less than...