A random sample of size 36 is taken from a population with mean [tex]$\mu=27$[/tex] and standard deviation [tex]$\sigma=6$[/tex]. The probability that the sample mean is greater than 30 is:
A. 0.0013
B. 0.6839
C. 0.2751
D. 0.4101
Answer (1)
Calculate the standard error: SE = n σ = 1 .
Calculate the z-score: z = SE x ˉ − μ = 3 .
Find the probability 3) = 1 - P(Z \le 3) \approx 1 - 0.99865 = 0.00135"> P ( Z > 3 ) = 1 − P ( Z ≤ 3 ) ≈ 1 − 0.99865 = 0.00135 .
The probability that the sample mean is greater than 30 is approximately 0.0013 .
Explanation
Understand the problem and provided data We are given a population with mean μ = 27 and standard deviation σ = 6 . We take a random sample of size n = 36 and want to find the probability that the sample mean is greater than 30.
Calculate the Standard Error First, we need to calculate the standard error of the mean (SE), which is the standard deviation of the sample mean. The formula for SE is: SE = n σ = 36 6 = 6 6 = 1
Calculate the Z-score Next, we calculate the z-score, which tells us how many standard errors the sample mean is away from the population mean. The formula for the z-score is: z = SE x ˉ − μ = 1 30 − 27 = 3
Find the Probability Now, we want to find the probability that the sample mean is greater than 30, which is equivalent to finding the probability that 3"> Z > 3 , where Z is a standard normal random variable. This is the same as 1 − P ( Z ≤ 3 ) .
Using a standard normal table or calculator, we find that P ( Z ≤ 3 ) ≈ 0.99865 . Therefore, 3) = 1 - P(Z \le 3) = 1 - 0.99865 = 0.00135"> P ( Z > 3 ) = 1 − P ( Z ≤ 3 ) = 1 − 0.99865 = 0.00135
Final Answer The probability that the sample mean is greater than 30 is approximately 0.00135. Comparing this to the given options, the closest answer is 0.0013.
Examples
Consider a scenario where you're analyzing the average test scores of students in a large school district. If you know the overall average score and standard deviation for all students, you can use this type of probability calculation to determine the likelihood that a random sample of students will have an average score significantly higher than the district average. This helps in identifying potentially high-performing groups or evaluating the effectiveness of specific educational programs. For example, if the district average is 27 with a standard deviation of 6, and you sample 36 students, you can calculate the probability of their average score exceeding 30 using the z-score and standard error formulas as demonstrated. This method provides valuable insights into sample performance relative to the population.
