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 [tex]\sigma = 1.93[/tex]
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.
Answer:
The likelihood is [tex]P(X \le 2.10 ) = 0.08931[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 55[/tex]
The population mean is [tex]\mu = 2.45 \ hours[/tex]
The random mean considered [tex]\= x = 2.10 \ hours[/tex]
Generally the standard error of mean is mathematically represented as
[tex]\sigma _{\= x } = \frac{\sigma }{ \sqrt{n} }[/tex]
=> [tex]\sigma _{\= x } = \frac{1.93 }{ \sqrt{55} }[/tex]
=> [tex]\sigma _{\= x } = 0.2602[/tex]
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
[tex]P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 1 - P( \frac{X - \mu }{\sigma_{\= x }} > \frac{ 2.10 - 2.45}{ 0.2602} )[/tex]
Generally [tex]\frac{X - \mu }{\sigma_{\= x }} = Z(The \ standardized \ value \ of \ X )[/tex]
[tex]P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 1 - P( Z > -1.345 )[/tex]
From the z-table the value of
[tex]P( Z > -1.345 ) = 0.91069[/tex]
So
[tex]P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 1 - 0.91069[/tex]
[tex]P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 0.08931[/tex]
[tex]P(X \le 2.10 ) = 0.08931[/tex]
