a) The sampling distribution for the sample mean would be a normal distribution with a mean equal to the population mean (12) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (4 / sqrt(9) = 2).
So, the mean of the sampling distribution would be 12, the standard deviation would be 2, and the shape would be normal.
b) We can use a standard normal table or a normal calculator function to find the probability of a person having moved at most 5 times. We need to first standardize the random variable by subtracting the population mean (12) and dividing by the population standard deviation (4):
z = (5 - 12) / 4 = -2.5
Using a standard normal table or a normal calculator function, we find that the probability of a person having moved at most 5 times is 0.0062.
c) To find the value such that 85% of the people will have moved at least that many times, we need to find the 85th percentile of the normal distribution. We can use a normal calculator function to find the inverse standard normal value for 0.85. Let x be the number of moves.
z = (x - 12) / 4
x = 12 + 4 * invNorm(0.85)
Using a normal calculator function, we find that invNorm(0.85) = 0.841. So,
x = 12 + 4 * 0.841 = 16.164
Rounding to the nearest whole number, we find that 85% of people from the United States will have moved at least 16 times in their lifetime.
d) To find the probability that a random sample of 9 people will have a mean number of moves greater than 17, we need to standardize the sample mean:
z = (17 - 12) / (2) = 2.5
Using a standard normal table or a normal calculator function, we find that the probability of a standard normal random variable being greater than 2.5 is 0.0062.
So, the probability that a random sample of 9 people will have a mean number of moves greater than 17 is 0.0062.
e) To find the probability that a random sample of 9 people will have a mean number of moves less than 9, we need to standardize the sample mean:
z = (9 - 12) / (2) = -1.5
Using a standard normal table or a normal calculator function, we find that the probability of a standard normal random variable being less than -1.5 is 0.0668.
So, the probability that a random sample of 9 people will have a mean number of moves less than 9 is 0.0668.
f) To find the middle 50% of sample means, we need to find the 25th and 75th percentiles of the normal distribution of sample means. We can use a normal calculator function to find the inverse standard normal values for 0.25 and 0.75. Let x be the number of moves.
z = (x - 12) / (2)
x = 12 + 2 * invNorm(0.25) or x = 12 + 2 * invNorm(0.75)
Using a normal calculator function, we find that invNorm(0.25) = -0.675 and invNorm(0.75) = 0.675. So,
x = 12 + 2 * -0.675 = 11.35 or x = 12 + 2 * 0.675 = 13.