Final answer:
To find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4, we can use the Normal Distribution.
Step-by-step explanation:
To find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4, we can use the Normal Distribution. Since the population has a discrete uniform distribution, the distribution of the sample means will be approximately normal. The mean of the sample means distribution will be the same as the population mean, which is 4. The standard deviation of the sample means distribution can be calculated using the formula:
Standard Deviation of Sample Means = Population Standard Deviation/sqrt (Sample Size)
So, in this case, the standard deviation of the sample means distribution is:
Standard Deviation of Sample Means = 0.2 / sqrt(54)
Now, we can use the standard deviation of the sample means distribution to calculate the probability that a sample mean is between 4.1 and 4.4 using the Normal Distribution. We can subtract the cumulative probability of 4.1 from the cumulative probability of 4.4 to get the desired probability.
For example, using a statistics calculator or table, we find that the cumulative probability of a Z-score of -0.5 (corresponding to 4.1) is 0.3085 and the cumulative probability of a Z-score of -0.3 (corresponding to 4.4) is 0.3821. So the probability that a random sample of size 54 will yield a sample mean greater than 4.1 but less than 4.4 is approximately 0.3821 - 0.3085 = 0.0736.