Question

The weights of medium oranges packaged by an orchard are Normally distributed with a mean of 14 ounces and a standard deviation of 2 ounces. Ten medium oranges will be randomly selected from a package. What is the sampling distribution of the sample mean weight of a random sample of 10 medium oranges

Answer 1

The sampling distribution of the sample mean weight of 10 medium oranges is
`\bar X` ~ Normal (
`\mu=14 \text{ ounces},s = 0.63 \text{ ounces}`).

Explanation:

We are given that the weights of medium oranges packaged by an orchard are Normally distributed with a mean of 14 ounces and a standard deviation of 2 ounces.

Ten medium oranges will be randomly selected from a package.

Let
`\bar X` = sample mean weight

The z-score probability distribution for sample mean is given by;

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean weight = 14 ounces

`\sigma` = population standard deviation = 2 ounces

n = sample of oranges selected = 10

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, The sampling distribution of sample mean weight (
`\bar X`) is given by;

`\bar X` ~ Normal (
`\mu =14, s = (\sigma)/(√(n) ) =(2)/(√(10) )` )

So,
`\bar X` ~ Normal (
`\mu=14 \text{ ounces},s = 0.63 \text{ ounces}`)