A study found that citizens spend on average $1950 per year on groceries with a standard deviation of $100. Assume that the variable is normally distributed.
5. Identify the population mean.
6. Identify the population standard deviation.
7. Find the probability that a sample of 20 citizens will have a mean less than $1800.
8. Find the probability that a sample of 500 citizens will have a mean greater than $2000.
Don't know population mean and SD, but study data is best estimate.
5. $1950
6. $100
7. Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
8. Same process as 7.
5. To identify the population mean, we need to refer to the information given in the question. In this case, the study found that citizens spend on average $1950 per year on groceries. Therefore, the population mean is $1950.
6. Similarly, to identify the population standard deviation, we need to refer to the information given in the question. The study states that the standard deviation is $100. Therefore, the population standard deviation is $100.
7. To find the probability that a sample mean is less than $1800, we can use the z-score formula and the standard normal distribution. The formula to calculate the z-score is:
z = (x - μ) / (σ / √n)
where:
z = z-score
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size
In this case, we want to find the probability for a sample of 20 citizens, so n = 20. We have x = $1800, μ = $1950, and σ = $100. Plugging these values into the formula, we get:
z = ($1800 - $1950) / ($100 / √20) ≈ -3.16
Now, we can use a standard normal distribution table or a calculator to determine the probability associated with a z-score of -3.16. Let's assume the probability is P(z < -3.16).
8. To find the probability that a sample mean is greater than $2000, we once again use the z-score formula and the concept of the standard normal distribution. Following a similar procedure as in question 7, we substitute x = $2000, μ = $1950, σ = $100, and the sample size n, which is now 500, into the formula:
z = ($2000 - $1950) / ($100 / √500) ≈ 10
Using the standard normal distribution table or a calculator, we can find the probability associated with a z-score of 10. Let's assume the probability is P(z > 10).