Statistics Homework Solutions
Problem
#4465

Determining the sampling distribution of a proportion.

According to a well-known newspaper journal, 58.3% of the 222,900 households in a particular county get the Sunday edition of the local newspaper.  Suppose that random samples of 200 households are selected.

In what proportion of the samples will between 55% and 60% of the households get the Sunday edition of the newspaper?

P(0.55 < Ps <0.60) = P(-0.95 < Z < 0.49) = 0.5168

0.49 in the Cumulative Standardized Normal Distribution Table (= 0.6879)

What are the specific steps involved in converting P(0.55 < Ps <0.60) to P(-0.95 < Z < 0.49)?


Solution
What is this?
By OTA - Overall OTA Rating
Purchase Cost Now
$2.19 CAD (was ~$15.96)
Included in Download
  • Plain text response
$2.19 Instant Download
Add to Cart
Why you can trust BrainMass.com
  • Your Information is Secure
  • Best Online Academic Help Service
  • Students find real academic Success
Related Solutions
  • Distribution on a Standardized Aptitude Test - The distribution of scores on a standardized aptitude test is approximately normal with a mean of 500 and a standard deviation of 105. What is the minimum score needed to be in the top 5% on this test ...
  • Sampling Distribution of a Proportion - 6.30 Given a population in which the proportion of items with a desired attribute is p=0.25, if a sample of 400 is taken: a. What is the standard deviation of the sampling distribution of _ ...
  • Sample proportion - 40% of households prefer new package. The probability that a random sample of 300 households will have a sample proportion greater than .45 is: ______ The probability that the 300 household samp ...
  • What is the difference between a mean and a proportion? - What is the difference between a mean and a proportion?
  • T4-10 - Approximately 5% of U.S. families have a net worth in excess of $1 million and thus can be called Millionaires. However, a survey in the year 2000 found that 30% of Microsofts 31,000 employees were m ...
Browse