One consequence of the popularity of the internet is that it is thoughout to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.10 hours watching television on a weekday. Required:
Determine the likelihood of obtaining a sample mean of 2.10 hours or less from a population whose mean is presumed to be 2.45 hours.

Complete  Question

One consequence of the popularity of the internet is that it is throughout to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.10 hours watching television on a weekday. The  standard deviation  is  

Required:

Determine the likelihood of obtaining a sample mean of 2.10 hours or less from a population whose mean is presumed to be 2.45 hours.

The  likelihood is  

Step-by-step explanation:

From the question we are told that  

    The sample size is  

    The population mean is  

    The random mean considered

Generally the standard error of  mean is mathematically represented as

=>          

=>          

The  likelihood of obtaining a sample mean of 2.10 hours or less from a population whose mean is presumed to be 2.45 hours is mathematically represented as

Generally    

From the z-table the value of

  So

answer: r there any options?

step-by-step explanation:

hi there!

the answer will be d: 2