Let Y = the number of hours a student studies for the midterm.
From previous semesters we know that Y ∼ N(6, 1.5). What is the probability that for 100 randomly chosen students, their mean study time will be more than 6.3 hours?
To find the probability that the mean study time for 100 randomly chosen students will be more than 6.3 hours, we can use the Central Limit Theorem. According to the Central Limit Theorem, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution, as the sample size increases.
In this case, we are given that the individual study time Y follows a normal distribution with a mean of 6 and a standard deviation of √1.5. The standard deviation of the sample mean, also known as the standard error (SE), can be calculated by dividing the population standard deviation by the square root of the sample size. Since we are dealing with 100 randomly chosen students, the sample size is 100. Therefore, the standard error can be calculated as follows:
SE = √(Population Standard Deviation / Sample Size)
= √(1.5 / 100)
= √0.015
= 0.1225
Now, we can standardize the value of 6.3 using the Z-score formula:
Z = (X - μ) / SE
Where X is the value we want to find the probability for (6.3 hours in this case), μ is the population mean (6), and SE is the standard error (0.1225).
Z = (6.3 - 6) / 0.1225
= 2.44898
To find the probability that the sample mean will be more than 6.3 hours, we need to find the area under the standard normal distribution curve to the right of Z = 2.44898.
Using a Z-table or a statistical software, we can find that the probability corresponding to this Z-score is approximately 0.0073.
Therefore, the probability that for 100 randomly chosen students, their mean study time will be more than 6.3 hours is approximately 0.0073, or 0.73%.