Probability of people having answering machines

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SUMMARY

The discussion centers on the probability of telephone subscribers having answering machines, with a survey indicating that 70% of subscribers possess one. When contacting 200 subscribers, the probability of 150 or more having an answering machine can be calculated using binomial distribution. The expected value (mean) of subscribers with answering machines is 140, and the standard deviation can be derived from the binomial parameters. For simplification, the normal distribution approximation can be applied, incorporating integer correction for accuracy.

PREREQUISITES
  • Understanding of binomial distribution and its parameters
  • Familiarity with normal distribution and its approximation techniques
  • Knowledge of expected value and standard deviation calculations
  • Basic probability concepts and tree diagrams
NEXT STEPS
  • Learn about binomial distribution calculations and applications
  • Study normal distribution approximation methods for binomial scenarios
  • Explore expected value and standard deviation in probability theory
  • Review integer correction techniques in statistical analysis
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Students and professionals in statistics, data analysts, and anyone involved in probability calculations related to telemarketing or survey analysis.

Larrytsai
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According to a research company survey, 70% of telephone subscrivers have an answering machine. If a telemarketing campaign contacts 200 telephone subscribers, what is the probability that 150 or more iwll have an answering machine?

Im lost for the tree diagram but and i have no clue for what method but my attempt was 0.7x 200 = 140 then 140/150, 1-(140/150)
 
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Have you learned about binomial distribution yet?
Then can you argue that the number of subscribers that has an answering machine is binomially distributed?
 


Of 200 people, 70% of whom, have answering machines, what is the "expected value" (mean) of the number who have answering machines? What is the standard deviation?

And if the number, 200, makes the calculations too tedious, you could use the "normal distribution approximation"- the normal distribution having the same mean and standard deviation. Don't forget to use the "integer correction". Since the normal distribution is a continuous distribution while the binomial is discrete, 149.5 would be rounded to 150.
 

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