The mean television viewing time for Americans is 15 hours per week. Suppose a sample of 60 Americans is taken to further investigate viewing habits. Assume the population standard deviation for weekly viewing ime is o = 4 hours.
A What is the probability the sample mean will be within 1 hour of the population mean?
B. What is the probability the sample mean will be within 45 minutes of the population mean?
You need to convert the hours to a Z score and then look up the probabilities in the back of your statistics textbook in an appendix labeled something like "areas under the normal distribution."
Z = (X - ƒÊ)/SD, where X = raw score, ƒÊ = mean and SD = standard deviation.
A involves X of 14 to 16, while B involves X of 14.25 to 15.75, converting minutes to portions of an hour.
You should be able to solve it with this information. I hope it helps. Thanks for asking.
Just wanted to say thank you for your help. It is greatly appreciated.
kjjn
0.38680
NO IDEA PLZ HELP ME
To find the probability in both scenarios, we need to use the concept of the sampling distribution and the Z-score.
The sampling distribution of the sample mean follows a normal distribution with a mean equal to the population mean (μ) and a standard deviation equal to the population standard deviation (σ) divided by the square root of the sample size (n).
For this problem:
Population mean (μ) = 15 hours per week
Population standard deviation (σ) = 4 hours
Sample size (n) = 60
A. What is the probability the sample mean will be within 1 hour of the population mean?
Step 1: Calculate the standard deviation of the sampling distribution (also known as the standard error):
Standard error (SE) = σ / sqrt(n)
SE = 4 / sqrt(60) = 0.5164 (rounded to 4 decimal places)
Step 2: Calculate the Z-score for the given deviation of 1 hour using the formula:
Z = (X - μ) / SE
Z = (1 - 15) / 0.5164 = -27.2587 (rounded to 4 decimal places)
Step 3: Find the probability of the sample mean being within 1 hour of the population mean by looking up the Z-score in the standard normal distribution table (or using a calculator).
The probability associated with Z = -27.2587 is extremely close to 0 or effectively 0.
B. What is the probability the sample mean will be within 45 minutes of the population mean?
Step 1: Convert 45 minutes to hours:
45 minutes = 0.75 hours
Step 2: Calculate the Z-score for the given deviation of 0.75 hours:
Z = (X - μ) / SE
Z = (0.75 - 15) / 0.5164 = -28.9603 (rounded to 4 decimal places)
Step 3: Find the probability of the sample mean being within 45 minutes of the population mean by looking up the Z-score in the standard normal distribution table (or using a calculator).
The probability associated with Z = -28.9603 is effectively 0.
In both cases, the probability of the sample mean being within the given deviations of the population mean is practically zero due to the extreme Z-scores resulting from the large deviations and the large sample size.