Question With such a large number of people using text messages as a means of communication, a company is interested in determining the number of work hours lost due to text messaging. Based on a survey of 32 randomly selected employees (anonymously, of course), the company has determined that the average amount of time spent texting over a one-month period is 173 minutes with a standard deviation of 66 minutes. What is the probability that the average amount of time spent using text messages is more than 199 minutes in this one-month period? Round your answer to four decimal places.

JEV1IX The Asker · Probability and Statistics

Transcribed Image Text: With such a large number of people using text messages as a means of communication, a company is interested in determining the number of work hours lost due to text messaging. Based on a survey of 32 randomly selected employees (anonymously, of course), the company has determined that the average amount of time spent texting over a one-month period is 173 minutes with a standard deviation of 66 minutes. What is the probability that the average amount of time spent using text messages is more than 199 minutes in this one-month period? Round your answer to four decimal places.
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Transcribed Image Text: With such a large number of people using text messages as a means of communication, a company is interested in determining the number of work hours lost due to text messaging. Based on a survey of 32 randomly selected employees (anonymously, of course), the company has determined that the average amount of time spent texting over a one-month period is 173 minutes with a standard deviation of 66 minutes. What is the probability that the average amount of time spent using text messages is more than 199 minutes in this one-month period? Round your answer to four decimal places.
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ASPSAS

Formulas:If mu,sigma are the mean and standard deviation of the population. If we take n samples out of the population then sample mean is mu_( bar(X))=mu and sample standard deviation is sigma_( bar(X))=(sigma)/(sqrtn)Facts :If X,Z be the normal and standard normal distibuted random variables respectively.To find the probabilites for X we convert X to Z by using Z=(X-mu)/(sigma)Where mu=mean(X)sigma= standard deviation (X)Let X= The time (in minutes) spent texting in one-month period.Given thatX follows an unknown distribution with mean (mu)=173and standard deviation (sigma)=66.bar(X) is sample mean(Sample average) ... See the full answer