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

The sampling distribution of the sample mean weight of 10 medium oranges is ~ Normal ().

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 = sample mean weight

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

                           Z = ~ N(0,1)

where, = population mean weight = 14 ounces

            = 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 () is given by;

              ~ Normal ( )

So,   ~ Normal ()

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation: