Final answer:
The probability of obtaining a sample mean less than 19.385 from a population mean of 20 and a standard deviation of 4, with a sample size of 64, is approximately 10.93% using the normal distribution.
Step-by-step explanation:
To calculate the probability that we will obtain a sample mean less than 19.385, given a population mean (μ) of 20 and a population standard deviation (σ) of 4, with a sample size (n) of 64, we can use the Central Limit Theorem (CLT). According to the CLT, the sampling distribution of the sample mean is normally distributed if the sample size is large enough, which is generally considered the case for n ≥ 30. In this case, the sample size is 64, so we can assume that the sampling distribution of the sample mean is normal.
First, we need to calculate the standard error of the mean (SEM), which is σ divided by the square root of n:
SEM = σ / √n
SEM = 4 / √64
SEM = 4 / 8
SEM = 0.5
Now, we find the z-score for the sample mean of 19.385:
z = (x - μ) / SEM
z = (19.385 - 20) / 0.5
z = -0.615 / 0.5
z = -1.23
Finally, using a z-score table or a statistical software, we can find the probability that Z is less than -1.23. This is the area under the normal curve to the left of the z-score -1.23, which we can determine is approximately 0.1093. This means there is an approximately 10.93% chance that the sample mean will be less than 19.385 when drawn from this population.