In Exercises 55–60, find the indicated probabilities and interpret the results.

The mean MCAT total score in a recent year is 500.9. A random sample of 32 MCAT total scores is selected. What is the probability that the mean score for the sample is (a) less than 503? Assume sigma=10.6.

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Step 1: Identify the given values in the problem. The population mean (μ) is 500.9, the population standard deviation (σ) is 10.6, the sample size (n) is 32, and we are tasked with finding the probability that the sample mean (x̄) is less than 503.

Step 2: Recall that the sampling distribution of the sample mean follows a normal distribution if the population is normal or the sample size is sufficiently large (n ≥ 30). The mean of the sampling distribution is μ, and the standard error (SE) is calculated as SE = σ / √n.

Step 3: Calculate the standard error (SE) using the formula SE = σ / √n. Substitute the given values: σ = 10.6 and n = 32. This will give you the standard deviation of the sampling distribution.

Step 4: Convert the sample mean (503) to a z-score using the formula z = (x̄ - μ) / SE. Substitute the values x̄ = 503, μ = 500.9, and the SE calculated in the previous step.

Step 5: Use the z-score obtained in Step 4 to find the cumulative probability from the standard normal distribution table or a statistical software. This cumulative probability represents the probability that the sample mean is less than 503.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Central Limit Theorem

The Central Limit Theorem states that the distribution of the sample mean will approach a normal distribution as the sample size increases, regardless of the population's distribution, provided the sample size is sufficiently large (typically n > 30). This theorem is crucial for calculating probabilities related to sample means, especially when dealing with a known population mean and standard deviation.

Standard Error of the Mean

The Standard Error of the Mean (SEM) quantifies how much the sample mean is expected to vary from the true population mean. It is calculated by dividing the population standard deviation (sigma) by the square root of the sample size (n). In this case, SEM helps determine the probability of the sample mean being less than a specific value, such as 503.

Z-Score

A Z-score measures how many standard deviations an element is from the mean. In the context of this problem, the Z-score is used to standardize the sample mean to find the corresponding probability in the standard normal distribution. By calculating the Z-score for the sample mean of 503, we can determine the likelihood of obtaining a sample mean less than this value.

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