I am working on a review sheet and got everything except for this one. I am not sure whether to use z or t.
The number of hours per week that high school juniors watch TV is normally distributed with a mean of 8 hours and a standard deviation of 2 hours. If 100 students are chosen at random, find the probability that the mean for that sample is between 8.2 and 8.8.
a. 1598 b. .1586 c. .1156 d. .1152
Note: standard deviation/ the square root of n.
Check out the Central Limit Theorem
Use z-scores.
For this problem:
z = (x - mean)/(sd/√n)
With your data:
z = (8.2 - 8)/(2/√100) = 1.00
z = (8.8 - 8)/(2/√100) = 4.00
Answer: .1586 is the probability (check a z-table between z = 1.00 and z = 4.00)
Thanks!! That really helped. I had the steps you had, but was stuck on z vs t.
To solve this problem, we need to use the Central Limit Theorem, which states that the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough (typically, n ≥ 30).
In this case, we are given the mean (μ) and standard deviation (σ) of the population: μ = 8 hours and σ = 2 hours. We are also told that we are taking a sample of 100 students.
First, we need to calculate the standard deviation of the sample mean. According to the formula provided: standard deviation/ the square root of n, the standard deviation of the sample mean (σx̄) is calculated by dividing the population standard deviation by the square root of the sample size.
σx̄ = σ / √n = 2 / √100 = 2 / 10 = 0.2
Now we can use the z-score formula to find the probability that the sample mean is between 8.2 and 8.8. The formula for calculating the z-score is:
z = (x - μ) / σ
where x is the sample mean we want to calculate the probability for, μ is the population mean, and σ is the standard deviation of the sample mean.
For the lower limit of 8.2, the z-score is:
z1 = (8.2 - 8) / 0.2 = 0.2 / 0.2 = 1
For the upper limit of 8.8, the z-score is:
z2 = (8.8 - 8) / 0.2 = 0.8 / 0.2 = 4
Now we can use a z-table or a calculator to find the area under the standard normal distribution curve between these two z-scores.
Using a z-table, we can find the area between z = 1 and z = 4, which corresponds to the probability that the sample mean is between 8.2 and 8.8.
From the z-table, we find that the area between z = 1 and z = 4 is approximately 0.1586.
Therefore, the answer is option b. 0.1586.