A recent university survey on the monthly number of hours that students spend commuting states, the average monthly commute time is 20 hours.
If the standard deviation is 7 hours and a random sample of 49 students is selected.
a) What is the probability that the sample average is more than 22 hours?
b) What is the probability that the sample average is less than 21 hours?
I can't seem to figure out what distribution this question is? Would it be normal, binomial, poisson or something else?
You will have to use either tables in your textbook or some kind of software that deals with the Normal Distribution curve.
This is an excellent webpage for this question:
http://davidmlane.com/normal.html
In case your computer breaks down. For normal distribution:
Z = (score-mean)/SE = (22-20)/(7/√49) = ?
SEm = SD/√n
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability.
To solve this problem, we need to determine the distribution of the sample average. In this case, we can use the Central Limit Theorem (CLT) since the sample size is sufficiently large (n=49) and the population standard deviation is known.
The Central Limit Theorem states that the sampling distribution of the sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is sufficiently large.
In this case, the population average is given as 20 hours, and the population standard deviation is 7 hours. The sample size is 49.
a) To find the probability that the sample average is more than 22 hours, we can use the z-score formula and standardize the sample mean.
Step 1: Calculate the standard error of the sample mean (σx̄) using the formula:
σx̄ = σ / √n
where σ is the population standard deviation and n is the sample size.
σx̄ = 7 / √49 = 7 / 7 = 1
Step 2: Calculate the z-score using the formula:
z = (x - μ) / σx̄
where x is the value we want to find the probability for, μ is the population mean, and σx̄ is the standard error of the sample mean.
z = (22 - 20) / 1 = 2
Step 3: Look up the z-score in the standard normal distribution table or use a calculator to find the corresponding probability.
P(z > 2) = 1 - P(z ≤ 2)
Using the standard normal distribution table, we find that P(z ≤ 2) is approximately 0.9772.
Therefore, P(z > 2) = 1 - 0.9772 = 0.0228, or approximately 2.28%.
So, the probability that the sample average is more than 22 hours is approximately 2.28%.
b) To find the probability that the sample average is less than 21 hours, we can follow a similar process:
Step 1: Calculate the standard error of the sample mean (σx̄) as previously calculated: 1
Step 2: Calculate the z-score:
z = (x - μ) / σx̄
z = (21 - 20) / 1 = 1
Step 3: Look up the z-score in the standard normal distribution table or use a calculator to find the corresponding probability.
P(z < 1) is the same as P(z ≤ 1), which is approximately 0.8413.
Therefore, the probability that the sample average is less than 21 hours is approximately 0.8413, or 84.13%.
In summary, the distribution used in this problem is the normal distribution, and we used the Central Limit Theorem to approximate the sampling distribution of the sample means.