The amount of time the university professors devote to their jobs per week is normally dis- tributed with a mean of 52 hours and a standard deviation of 8 hours.
(a) What is the probability that a professor works for more than 56 hours per week?
(b) Find the probability that the mean amount of work per week for four randomly selected professors is more than 56 hours.
(c) Find the probability that the mean amount of work per week for 36 randomly selected professors is more than 56 hours.
(d) If the devoted time to job is not normally distributed, which, if any, of the questions can you answer? Explain.
To solve these questions, we will use the properties of the normal distribution. Let's calculate the probabilities step by step:
(a) What is the probability that a professor works for more than 56 hours per week?
To solve this, we need to find the area under the normal distribution curve to the right of 56 hours. We can use a z-score to determine this probability.
First, we calculate the z-score using the formula:
z = (x - mean) / standard deviation
In this case, x = 56 hours, mean = 52 hours, and standard deviation = 8 hours.
z = (56 - 52) / 8 = 0.5
Using a z-score table or a calculator, we find that the probability corresponding to a z-score of 0.5 is approximately 0.6915.
Therefore, the probability that a professor works for more than 56 hours per week is 0.6915 or 69.15%.
(b) Find the probability that the mean amount of work per week for four randomly selected professors is more than 56 hours.
Since the sample size is larger than 30, we can use the Central Limit Theorem. According to this theorem, the distribution of the sample means approaches a normal distribution as the sample size increases.
To find the probability, we need to calculate the z-score for the sample mean using the formula:
z = (x̄ - mean) / (standard deviation / √n)
In this case, x̄ = 56 hours, mean = 52 hours, standard deviation = 8 hours, and n = 4 professors.
z = (56 - 52) / (8 / √4) = 2
Using a standard normal distribution table or a calculator, the probability corresponding to a z-score of 2 is approximately 0.9772.
Therefore, the probability that the mean amount of work per week for four randomly selected professors is more than 56 hours is 0.9772 or 97.72%.
(c) Find the probability that the mean amount of work per week for 36 randomly selected professors is more than 56 hours.
Using the same formula as in part (b) but with n = 36 professors:
z = (56 - 52) / (8 / √36) = 3
Using a standard normal distribution table or a calculator, the probability corresponding to a z-score of 3 is approximately 0.9987.
Therefore, the probability that the mean amount of work per week for 36 randomly selected professors is more than 56 hours is 0.9987 or 99.87%.
(d) If the devoted time to the job is not normally distributed, which, if any, of the questions can you answer? Explain.
If the devoted time to the job is not normally distributed, we would not be able to accurately answer any of the questions using the properties and assumptions of the normal distribution.
However, it is worth noting that if we have information about the alternative distribution, such as another known distribution or sufficient data to estimate the parameters of a different distribution, we might be able to solve the questions using those assumptions instead.
(a) To find the probability that a professor works for more than 56 hours per week, we need to standardize the value using the z-score formula and then use the standard normal distribution table.
The z-score formula is given by: z = (x - μ) / σ
where x is the value we are interested in (56 hours in this case), μ is the mean (52 hours), and σ is the standard deviation (8 hours).
Let's calculate the z-score:
z = (56 - 52) / 8
z = 4 / 8
z = 0.5
Using the standard normal distribution table, we can find the probability associated with the z-score of 0.5.
The probability that a professor works for more than 56 hours per week is equal to the area under the standard normal curve to the right of the z-score of 0.5.
Looking up the z-score of 0.5 in the standard normal distribution table, we find the corresponding cumulative probability to be approximately 0.69146.
Therefore, the probability that a professor works for more than 56 hours per week is approximately 0.69146, or 69.15%.
(b) To find the probability that the mean amount of work per week for four randomly selected professors is more than 56 hours, we can use the Central Limit Theorem.
According to the Central Limit Theorem, as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the original population distribution.
Since the sample mean follows a normal distribution with the same mean (52 hours) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (8 / √4 = 4), we can standardize the value using the z-score formula.
Let's calculate the z-score:
z = (56 - 52) / 4
z = 1
Using the standard normal distribution table, we can find the probability associated with the z-score of 1.
The probability that the mean amount of work per week for four randomly selected professors is more than 56 hours is equal to the area under the standard normal curve to the right of the z-score of 1.
Looking up the z-score of 1 in the standard normal distribution table, we find the corresponding cumulative probability to be approximately 0.84134.
Therefore, the probability that the mean amount of work per week for four randomly selected professors is more than 56 hours is approximately 0.84134, or 84.13%.
(c) Similar to part (b), we can use the Central Limit Theorem to find the probability that the mean amount of work per week for 36 randomly selected professors is more than 56 hours.
Since the sample mean follows a normal distribution with the same mean (52 hours) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (8 / √36 = 8/6 = 4/3), we can standardize the value using the z-score formula.
Let's calculate the z-score:
z = (56 - 52) / (4/3)
z = 4 / (4/3)
z = 4 * (3/4)
z = 3
Using the standard normal distribution table, we can find the probability associated with the z-score of 3.
The probability that the mean amount of work per week for 36 randomly selected professors is more than 56 hours is equal to the area under the standard normal curve to the right of the z-score of 3.
Looking up the z-score of 3 in the standard normal distribution table, we find the corresponding cumulative probability to be approximately 0.99865.
Therefore, the probability that the mean amount of work per week for 36 randomly selected professors is more than 56 hours is approximately 0.99865, or 99.865%.
(d) If the devoted time to the job is not normally distributed, we cannot answer any of the questions using the standard normal distribution and z-scores. In such cases, we would need to know the specific distribution of the data and use appropriate methods based on that distribution to answer the questions.
a) Z = (score-mean)/SD
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability related to the Z score.
b, c) Z = (score-mean)/SEm
SEm = SD/√n
Use same table.
I'll leave the explanation up to you.