Question

A greenhouse in a? tri-county area has kept track of its customers for the last several years and has determined that? 28% of its customers plant a vegetable garden in the spring. the greenhouse obtains a random sample of 1000 of its customers. what is the mean of the sampling distribution of modifyingabove p with caretp?, the sample proportion of customers that plant a vegetable garden in the? spring?

Answer 1

We are given two values
p = 28%
and
n = 1000
We are asked to get the mean of the sampling distribution of the customers that plant a vegetable in the spring.

We simply use the formula
p̂ = np
So, substituting
p̂ = 0.28 (1000)
p̂ = 280
The mean is 280

Answer 2

Answer:

The mean is 280 customers.

The sample proportion is 28%.

Explanation:

For each customer, there are only two possible outcomes. Either they plant a vegetable in the spring, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution.

Probability of exactly x sucesses on n repeated trials, with p probability. Has an expected value of:

$E(X) = np$

In this problem, we have that:

There are 1000 customers, so $n = 1000$.

28% of its customers plant a vegetable garden in the spring. This means that $p = 0.28$.

So:

What is the mean of the sampling distribution of modifyingabove p with caretp?

The mean is the expected value, so:

$E(x) = np = 1000*0.28 = 280$

The mean is 280 customers.

The sample proportion of customers that plant a vegetable garden in the? spring?

It is the same as the population proportion, so the sample proportion is 28%.