Suppose that the mass of Moderate Size Eastern Capybaras (MSEC's) is a random variable that is normally distributed with mean 60 kg and variance 25 kg squared. If a random sample of 100 MSECSs is selected, what is the probability that the average mass in that sample will exceed 60.8 kg? (You may assume that the Central Limit Theorem applies.)
Added by Steven G.
Close
Step 1
The population mean is $\mu = 60$ kg. The population variance is $\sigma^2 = 25$ kg squared. The population standard deviation is $\sigma = \sqrt{25} = 5$ kg. The sample size is $n = 100$. We are interested in the probability that the sample average mass, Show more…
Show all steps
Your feedback will help us improve your experience
Prabhakar Kumar and 100 other Intro Stats / AP Statistics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A sample of 100 northern koala bears is chosen. The mean of the sample will be found. a. Does the Central limit theorem apply to this situation? Why? b. Find the mean and the standard error for the sampling distribution of the weights of samples of 100 northern koala bears. Be sure to choose the correct notation and follow indicated rounding rules. Mean: μx̄ (if necessary round to one decimal place) Stand Error σx̄ if rounding is necessary round to three decimal places) c. What is the probability that if a sample of 100 northern koala bears is chosen that the mean weight will be greater than 15.3 pounds? Give your response to four decimal places.
Prabhakar K.
Central Limit Theorem with Means The mean weight of an adult is 60 kilograms with a variance of 100. If 118 adults are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 0.8 kilograms? Round your answer to four decimal places.
Narayan H.
5. A sample is taken from a population with mean 85 with standard deviation 15. Determine the probability that a random sample of size 277 will result in a sample mean of at least 86.3. Round your answer to three decimal places. known, n > 30 CLT: distribution of sample means approximately normal 1 - Φ ( (86.3 - 85) / (15 / √277) ) = 1 - Φ(1.44) = 1 - 0.9251 = 0.075
Ahmet Y.
Recommended Textbooks
Elementary Statistics a Step by Step Approach
The Practice of Statistics for AP
Introductory Statistics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD