Question

Finding Probability of Mean for Random Sample

The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. If the benefits are normally distributed with a standard deviation of$460, find the probability that the mean benefit for a random sample of 20 patients is Less than $3800
2 0
Add a comment Improve this question Transcribed image text
Answer #1
The probability is
P(xbar<3800) = P((xbar-mean)/(s/vn) <(3800-4064 )/(460/sqrt(20)))
=P(Z<-2.57)
= 0.0051 (check standard normal table)
Add a comment
Answer #2

μ=4064

σ=460

n=20

xbar = sample mean

P(xbar < 3800) = P( (xbar-μ)/(σ/√n) < (3800-4064)/(460/√20) ) = P(z < -2.57) = .0051

answered by: Bluebird
Add a comment
Know the answer?
Add Answer to:
Finding Probability of Mean for Random Sample
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Medicare Hospital Insurance The average yearly Medicare Hospital Insurance benefit per person was $4064 in a...

    Medicare Hospital Insurance The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. Suppose the benefits are normally distributed with a standard deviation of $460. Round the final answers to at least four decimal places and intermediate z-value calculations to two decimal places. Part 1 of 2 Find the probability that the mean benefit for a random sample of 28 patients is less than $3870. p(x < 3870) - LE dl. Part 2 of 2...

  • The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. If...

    The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. If the benefits are normally distributed with a standard deviation of $460, find the probability that the mean benefit for a random sample of 20 patients is more than $4100. According to the U.S.Census, 67.5% of the U.S. population were born in their state of residence. In a random sample of 200 Americans, what is the probability that fewer than 125 were born in their...

  • Hey guys! Looking for some help with statistics homework. These online classes have made it a...

    Hey guys! Looking for some help with statistics homework. These online classes have made it a little hard to understand some of the specifc stuff like finding Z-score. 1) The average yearly Medicare Insurance benefit per person was $4064 in a recent year. The benefits are normally distributed with a standard deviation of $460 and 20 patients are randomly sampled: a) Find the mean and standard deviation of the sampling distribution of the sample mean: b) Find the probability that...

  • Find the probability that the mean of a random sample of 25 elements from a normally...

    Find the probability that the mean of a random sample of 25 elements from a normally distributed population with a mean of 90 and a standard deviation of 60 is larger than 100.

  • 6.33 Let x be a continuous random variable that is normally distributed with a mean of 25 and a standard deviation o...

    6.33 Let x be a continuous random variable that is normally distributed with a mean of 25 and a standard deviation of 6. Find the probability that x assumes a value a. between 28 and 34 b. between 20 and 35 6.34 Let x be a continuous random variable that has a normal distribution with a mean of 30 and a stan- dard deviation of 2. Find the probability that x assumes a value a. between 29 and 35 b....

  • In a hospital, a random sample of 10 weeks was selected, and it was found that...

    In a hospital, a random sample of 10 weeks was selected, and it was found that an average of 495 patients were treated in the emergency room each week. The population standard deviation was 23. Find the 90% confidence interval of the true mean. Assume the variable is normally distributed. (Show Work)

  • 2) In a hospital, a random sample of 12 weeks was selected, and it was found...

    2) In a hospital, a random sample of 12 weeks was selected, and it was found that an average sample of 537 patients were treated in the emergency room each week. The standard deviation was 21.82. Assuming that the variable is normally distributed, find the 90% confidence interval of the true mean. 90% confidence interval:

  • Answer and sketch questions 1 and 2 1. a) A random sample of size 36 is...

    Answer and sketch questions 1 and 2 1. a) A random sample of size 36 is selected from a normally distributed population with a mean of 16 and a standard deviation of 3. What is the probability that the sample mean is somewhere between 15.8 and 16.2? b). A normally distributed population has a mean of 16 and a standard deviation of 3. What is the probability that a single event is somewhere between 15.8 and 16.2? 2. A bank...

  • please the question. need R code to solve the question Assume that the literature on this...

    please the question. need R code to solve the question Assume that the literature on this topic suggests that the distribution of days of hospital stay are normally distributed with a mean of 5 and a standard deviation of 3. Recent publications have indicated that hypervirulent strains of C. Difficile are on the rise. Such strains are associated with poor outcomes, including extended hospital stays.An investigator is interested in showing that the average hospital stay durations have increased versus published...

  • the sat scores for males on the critical reading portion of the sat are normally distributed...

    the sat scores for males on the critical reading portion of the sat are normally distributed with a mean of 498 and standard deviation of 116. a. fund the probability that a randomly selected person scores higher than 700. b. Find the probability that a randomly selected person score's Less than 600. c. random samples of size n=20 are drawn from the population of male critical reading sat scores, And The mean of each sample is determined. use the central...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT