Respuesta :
Answer:
The sampling distribution of the sample mean weight of 10 medium oranges is [tex]\bar X[/tex] ~ Normal ([tex]\mu=14 \text{ ounces},s = 0.63 \text{ ounces}[/tex]).
Step-by-step 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 [tex]\bar X[/tex] = sample mean weight
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean weight = 14 ounces
[tex]\sigma[/tex] = 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 ([tex]\bar X[/tex]) is given by;
[tex]\bar X[/tex] ~ Normal ( [tex]\mu =14, s = \frac{\sigma}{\sqrt{n} } =\frac{2}{\sqrt{10} }[/tex] )
So, [tex]\bar X[/tex] ~ Normal ([tex]\mu=14 \text{ ounces},s = 0.63 \text{ ounces}[/tex])
