The weekly study times for all students in a school are normally distributed with μ = 18 hours and σ = 6 hours. If 36 students are randomly selected, find the probability that their mean study time is greater than 20 hours. My work: 20-18/(6/√36)=2/(6/6)=2/1=2 .4772 47.72% probability that their mean study time is greater than 20 hours. But answer should be 62.93.
Probability answer never > 1.
If Z = 2, P > 2 = .0228
To find the probability that the mean study time for a sample of 36 students is greater than 20 hours, you can use the Central Limit Theorem and the standard normal distribution.
The Central Limit Theorem states that if the sample size is large enough (typically greater than 30) and the population distribution is approximately normal, then the distribution of the sample means will be approximately normal, regardless of the shape of the population distribution.
First, let's calculate the standard deviation (σ) of the sample mean, which is also known as the standard error of the mean. Since the population standard deviation (σ) is given as 6 hours and the sample size (n) is 36, the standard error (SE) can be calculated as:
SE = σ / √n
SE = 6 / √36
SE = 6 / 6
SE = 1
Next, we compute the z-score using the formula:
z = (x - μ) / SE
For the given problem, we want to calculate the probability of the mean study time being greater than 20 hours. We can convert this into a z-score:
z = (20 - 18) / 1
z = 2
Finally, we look up the area (probability) to the right of the z-score of 2 in the standard normal distribution table. The probability of getting a z-score greater than 2 is 0.0228.
However, note that the z-score we calculated corresponds to the probability of a single student studying more than 20 hours, not the mean of a sample of 36 students. To find the probability of the mean study time for the sample being greater than 20 hours, we need to calculate the z-score of the sample mean.
Since the sample mean follows a normal distribution with the same mean (μ = 18) and a standard error (SE = 1) as calculated previously, we can use the same z-score of 2.
Now, we need to find the area to the right of this z-score of 2 in the standard normal distribution table. The probability of getting a z-score greater than 2 is approximately 0.0228.
However, we are interested in the probability that the mean study time is greater than 20 hours, which means we want the area to the left of the z-score of 2. To find this probability, we subtract the probability from 0.5 (because the standard normal distribution is symmetric):
P(mean > 20) = 0.5 - 0.0228
P(mean > 20) = 0.4772
Therefore, based on these calculations, the probability that the mean study time for a sample of 36 students is greater than 20 hours is approximately 0.4772 or 47.72%.
It seems there might have been an error in the calculations provided, which led to the discrepancy in the final result.