The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a standard deviation of 6 hours. Suppose we select a random sample of 144 current students. What is the probability that the mean of this sample is less than 19.25 hours?
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To solve this problem, we can use the Central Limit Theorem, which states that for a large enough sample size, the sampling distribution of the mean will be approximately normal regardless of the shape of the population distribution.
Given:
Population Mean (μ) = 20 hours
Population Standard Deviation (σ) = 6 hours
Sample Size (n) = 144
Sample Mean (x̄) < 19.25 hours
To find the probability, we need to standardize the sample mean using z-score formula:
z = (x̄ - μ) / (σ / √n)
Substituting the given values:
z = (19.25 - 20) / (6 / √144)
z = -0.75 / (6 / 12)
z = -0.75 / 0.5
z = -1.5
Now, we need to find the probability that the z-score is less than -1.5 using a standard normal distribution table or a calculator.
Looking up the z-score in the standard normal distribution table, we find that the probability corresponding to z = -1.5 is approximately 0.0668.
Therefore, the probability that the mean of this sample is less than 19.25 hours is approximately 0.0668 or 6.68%.
To find the probability that the mean of the sample is less than 19.25 hours, we need to use the concept of the sampling distribution of the sample mean.
The sampling distribution of the sample mean follows the normal distribution if the sample size is large enough, which is determined by the Central Limit Theorem. In this case, since the sample size is 144 (which is generally considered large), we can assume that the distribution of the sample mean will be approximately normal.
The mean of the sampling distribution of the sample mean is equal to the population mean, which is 20 hours in this case. The standard deviation of the sampling distribution of the sample mean (also known as the standard error) can be calculated using the formula:
standard error = standard deviation / sqrt(sample size)
In our case, the standard deviation is 6 hours and the sample size is 144. Therefore,
standard error = 6 / sqrt(144) = 6 / 12 = 0.5 hours.
Now, we can convert the question into a z-score problem. The z-score represents the number of standard deviations away from the mean a particular value is. In this case, we want to find the probability that the sample mean is less than 19.25 hours, so we calculate the z-score using:
z = (19.25 - mean) / standard error = (19.25 - 20) / 0.5 = -0.75.
Now, we need to find the probability that the z-score is less than -0.75. This can be done by looking up the value in the standard normal distribution table or using a calculator.
Using a standard normal distribution table or a calculator, the probability that the z-score is less than -0.75 is approximately 0.2266.
Therefore, the probability that the mean of the sample is less than 19.25 hours is approximately 0.2266 or 22.66%.