SOLUTION: 1. A survey reports its results by stating that standard error of the mean to be is 20. The population standard deviation is 500. (i). How large is the sample used in this survey?

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Question 1177196: 1. A survey reports its results by stating that standard error of the mean to be is 20. The population standard deviation is 500.
(i). How large is the sample used in this survey?
(ii). What is the probability that the sample mean will be within 25 of the population mean?
2(a). The time it takes a fire department to respond to a request for emergency aid has a mean of 14 minutes with standard deviation of 4 minutes. Suppose we randomly sample 50 emergency requests over a two-month period.
(i). State the sampling distribution of the sample mean, ̅x.
(ii). What is the probability that the sample mean will be 15 minutes or less?
(iii). What role does the Central Limit Theorem play in identifying this distribution?
(b). A population has a mean of 400 and a standard deviation of 50. The probability distribution of the population is unknown. A research study will use a simple random sample of either 10, 20, 35, or 50. items to collect data about the population.
(i). In which of these sample sizes will you be able to use the Normal probability distribution to describe the sampling distribution of the sample mean, ̅x. Explain.
(iii). Determine the standard error of the sampling distribution of ̅x for the instances where the Normal distribution would be appropriate.
(iii). What theorem would you appeal to for the justification of your answers in (i) and (ii). State it.
Answer by ewatrrr(24785) About Me  (Show Source):
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Hi

!.  SE+=+sigma%2Fsqrt%28n%29 = 20
i. n+=+%28sigma%2FSE%29%5E2 = (500/20)^2 = 625

ii. ME =+z%2Asigma%2Fsqrt%28n%29 =25
   25 = z*20
   z = 25/20 = 1.25  and P (z ≤  1.25) = .8944

2a.  z+=blue+%28x+-+14%29%2Fblue%284%2Fsqrt%2850%29%29 
       1/SE = z =  1.7678  and P(z ≤  1.7678) = .9615
The central limit theorem states that whenever a random sample of size n is taken from any distribution with mean and variance, then the sample mean will be approximately normally distributed with mean and variance
 We are proceeding as if the distribution of this sample(size 50)..its mean will be approximately normally distributed.

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