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Suppose the average fresh fruit consumption in the United States is 99.95 lbs. with a standard deviation of 30 lbs. Assume that fresh fruit consumption varies normally. A researcher took a simple random sample (SRS) of 50 people and had them record their fresh fruit consumption.

Answer the following questions:

i) Find the range of individual fruit consumption to be in the middle 99% area.

j) Write the sampling distribution of the sample mean fruit consumption and find the probability that a sample mean fruit consumption is below 94 lbs.

k) Find the range of average fruit consumption to be in the middle 90% area.

Answer :

We are given information about the average fresh fruit consumption in the United States and asked to find the range of individual fruit consumption to be in the middle 99% area, the sampling distribution of sample mean fruit consumption, and the range of average fruit consumption to be in the middle 90% area.

i) Range of Individual Fruit Consumption:

  1. To find the range of individual fruit consumption to be in the middle 99% area, we need to find the z-scores corresponding to the 0.005th and 0.995th percentiles of the standard normal distribution.
  2. Using a z-table, the z-score corresponding to the 0.005th percentile is approximately -2.58, and the z-score corresponding to the 0.995th percentile is approximately 2.58.
  3. Now, we can calculate the range of individual fruit consumption as follows: Range = (99.95 lbs - 2.58 * 30 lbs) to (99.95 lbs + 2.58 * 30 lbs), which is approximately 23.6 lbs to 176.3 lbs.

ii) Sampling Distribution of Sample Mean Fruit Consumption:

  1. The sampling distribution of the sample mean fruit consumption is approximately normal with a mean equal to the population mean (99.95 lbs) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (30 lbs / sqrt(50)).
  2. To find the probability that a sample mean fruit consumption is below 94 lbs, we need to find the z-score corresponding to this value using the same formula as before.
  3. Using the z-score, we can look up the corresponding area in the standard normal distribution table, which gives us the probability.

iii) Range of Average Fruit Consumption:

  1. To find the range of average fruit consumption to be in the middle 90% area, we need to find the z-scores corresponding to the 0.05th and 0.95th percentiles of the standard normal distribution.
  2. Using a z-table, the z-score corresponding to the 0.05th percentile is approximately -1.645, and the z-score corresponding to the 0.95th percentile is approximately 1.645.
  3. Now, we can calculate the range of average fruit consumption as follows: Range = (99.95 lbs - 1.645 * (30 lbs / sqrt(50))) to (99.95 lbs + 1.645 * (30 lbs / sqrt(50))), which is approximately 95.1 lbs to 104.8 lbs.

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