A specific study found that the average number of doctor visits per year for people over 55 is 8 with a standard deviation of 2. Assume that the variable is normally distributed.
1. Identify the population mean.
The population mean is 8.
2. Identify the population standard deviation.
The population standard deviation is 2.
3. Suppose a random sample of 15 people over 55 is selected. What is the probability that the sample mean is above 9?
4. Suppose a random sample of 100 people over 55 is selected. What is the probability that the sample mean will be below 7?
1 & 2 are right.
3. 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.
4. Use same process.
To answer both questions 3 and 4, we can use the concept of the sampling distribution of the sample mean. The sampling distribution of the sample mean follows a normal distribution if the sample size is large enough.
In both cases, we need to compute the z-score, which indicates how many standard deviations the sample mean is away from the population mean. The formula for the z-score is:
z = (x - μ) / (σ / √n)
Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size
For question 3:
n = 15, μ = 8, σ = 2, x = 9
Calculating the z-score:
z = (9 - 8) / (2 / √15) = 0.87
Now, we can look up the area under the standard normal distribution curve to the right of the z-score 0.87 (approximately 0.1949). However, since we are interested in the probability of the sample mean being above 9, we need to subtract this value from 1.
The probability that the sample mean is above 9 is approximately 1 - 0.1949 = 0.8051 or 80.51%.
For question 4:
n = 100, μ = 8, σ = 2, x = 7
Calculating the z-score:
z = (7 - 8) / (2 / √100) = -5
We can now look up the area under the standard normal distribution curve to the left of the z-score -5. Since it is such an extreme value, the probability is practically 0.
The probability that the sample mean is below 7 is practically 0.
To answer questions 3 and 4, we need to use the concept of the sampling distribution of the sample mean.
The first step is to calculate the standard deviation of the sampling distribution of the sample mean. This is also known as the standard error. The formula for the standard error (SE) is:
SE = population standard deviation / square root of sample size
For question 3:
SE = 2 / sqrt(15) ≈ 0.516
Now, we can use the concept of z-scores to find the probability. The z-score is defined as the number of standard errors away from the mean. The formula for z-score is:
z = (sample mean - population mean) / SE
For question 3:
z = (9 - 8) / 0.516 ≈ 1.94
Using a standard normal distribution table or a calculator, we can find the probability that the z-score is above 1.94. The probability being asked for is the area under the curve to the right of the z-score.
For question 4, we can follow the same steps but with a different sample size and population mean.
For question 4:
SE = 2 / sqrt(100) = 0.2
z = (7 - 8) / 0.2 = -5
The probability being asked for is the area under the curve to the left of the z-score (-5).