To find the probability that the sample mean would be less than $18850, we can use the central limit theorem and assume that the sample mean follows a normal distribution.
First, calculate the standard error of the mean:
Standard error = standard deviation / sqrt(sample size)
SE = 469 / sqrt(333) = 25.61
Next, calculate the Z-score:
Z = (X - mean) / SE
Z = (18850 - 18904) / 25.61 = -2.11
Using a standard normal distribution table or calculator, find the probability that Z is less than -2.11. This probability is approximately 0.0174.
Therefore, the probability that the sample mean would be less than $18850 if a sample of 333 persons is randomly selected is 0.0174, rounded to four decimal places.
The mean per capita income is 18,904
dollars per annum with a standard deviation of 469
dollars per annum.
What is the probability that the sample mean would be less than 18850
dollars if a sample of 333
persons is randomly selected? Round your answer to four decimal places.
1 answer