Final answer:
To find the probability of obtaining a sample mean between 9.6 and 9.7, we need to calculate the z-scores for those two values and find the difference between the cumulative probabilities under the standard normal curve.
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 9.6 but less than 9.7, we need to look at the sampling distribution of sample means. For a discrete uniform population, the sample mean at a particular sample size is equal to the mean of the population. In this case, the mean is 9. We also know that the standard deviation of the population can be calculated as the square root of the variance, which is ((14-4+1)^2 - 1)/12 ≈ 4.0825.
Since we are assuming the means are measured to the nearest tenth, we can convert the sample mean of 9.6 and 9.7 to the standard normal distribution using the formula z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. By calculating the z-scores for both 9.6 and 9.7, we can find the area under the standard normal curve between those two z-scores to get the probability.
Using a standard normal distribution table or a calculator, we find that the z-score for 9.6 is (9.6 - 9) / (4.0825 / sqrt(54)) ≈ 0.7627, and the z-score for 9.7 is (9.7 - 9) / (4.0825 / sqrt(54)) ≈ 1.0146.
The probability of obtaining a sample mean between 9.6 and 9.7 can be found by calculating the difference between the cumulative probabilities for the two z-scores, which is approximately 0.1357.