According to a local plant store, $28 \%$ of its customers plant a vegetable garden in spring.
Suppose a random sample of 1000 of its customers was asked if they plant a vegetable garden in spring. describe the sampling distribution of the proportion of customers that plant a vegetable garden in spring.
Shape:
Mean:
Standard Deviation (round to four decimals):
In a random sample of 1000 customers, what is the probability that at least $30 \%$ of customers plant a vegetable garden in spring? Round to four decimals.
Answer
Therefore, the shape of the sampling distribution is approximately normal due to the large sample size. The mean of the sampling distribution is 280. The standard deviation of the sampling distribution is approximately 14.1986. The probability that at least 30% of customers plant a vegetable garden in spring is approximately 0.0795.
Step 1 ：The problem is asking for the sampling distribution of the proportion of customers that plant a vegetable garden in spring. This is a binomial distribution problem, where the probability of success (planting a vegetable garden in spring) is 0.28.
Step 2 ：The mean of a binomial distribution is np, and the standard deviation is sqrt(np(1-p)), where n is the number of trials (1000 in this case) and p is the probability of success.
Step 3 ：Using these formulas, we find that the mean is \(1000 \times 0.28 = \boxed{280}\) and the standard deviation is \(\sqrt{1000 \times 0.28 \times (1-0.28)} = \boxed{14.1986}\).
Step 4 ：The second part of the question is asking for the probability that at least 30% of customers plant a vegetable garden in spring. This is equivalent to finding the probability that the proportion is greater than or equal to 0.30.
Step 5 ：This can be found by calculating the z-score of 0.30 and finding the area to the right of this z-score in the standard normal distribution. The z-score is \(\frac{0.30 - 0.28}{\sqrt{0.28 \times (1-0.28)/1000}} = 1.4086\).
Step 6 ：Using a standard normal distribution table or calculator, we find that the probability that the z-score is greater than or equal to 1.4086 is approximately \(\boxed{0.0795}\).
Step 7 ：Therefore, the shape of the sampling distribution is approximately normal due to the large sample size. The mean of the sampling distribution is 280. The standard deviation of the sampling distribution is approximately 14.1986. The probability that at least 30% of customers plant a vegetable garden in spring is approximately 0.0795.
