According to the U.S. Census Bureau, the mean household income in the United States in 2000 was $57,045 and the median household income was $42,148 (U.S. Census Bureau, "Money Income in the United States: 2000," www.census.gov, September 2001). The variability of household income is quite large, with the 10th percentile approximately equal to $111,600, and an overall standard deviation of approximately $25,000. Suppose random samples of 225 households were selected.
a) What proportion of the sample means would be below $55,000?
b) If random samples of size 20 were selected, can you use the methods discussed in this chapter to calculate the probabilities requested in (a)-(c)? Explain.
Yes. These probabilities are independent of sample size.
Solution Preview
(a) x = 55000, mu = 57045, sigma = 25000
z = (x - mu)/(sigma/sqrt n)
z = (55000 - ...
Solution Summary
A Complete, Neat and Step-by-step Solution is provided.
Probability: Random Sampling. ... I've included my explanations in the excel. Thank you for using Brainmass. The expert examines random sampling for probability. ...
... x exceeds 110 c) Find the probability that the ... strength, mean, standard deviation and probabilities of normally ... the quality of the paper, a random sample of 10 ...
... A random sample of 50 cans is selected, and the ... use Poisson distribution to calculate probabilities concerning the ... The probability that a success will occur is ...
... If you select a random sample of 25 sessions ... standard normal tables, we can find the probabilities as follows ... a. what is the probability that the sample mean is ...
... water from a supermarket prefer brand x. If a random sample of 425 such consumers is taken, what is the probability that between 35% and 38% prefer brand x? ...
...Probability in terms of percentage = 68.26% 4. Suppose you have drawn a simple random sample of 10 students from a college campus and recorded how many hours ...
... of defective disks among 20 samples. Then we know X is a random variable and follows binomial distribution, ie., X~b(20,0.05). So, the probability that no ...
... Page 4 2(a). Normal Probabilities. ...Probability for X > X Value 134 Z Value 0.52762432 P(X>134) 0.30. ... If we selected another random sample of 50 participants,. ...
