Question

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.

Answer 1

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
`\sigma = 1.93`

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
`P(X \le 2.10 ) = 0.08931`

Explanation:

From the question we are told that

The sample size is
`n = 55`

The population mean is
`\mu = 2.45 \ hours`

The random mean considered
`\= x = 2.10 \ hours`

Generally the standard error of mean is mathematically represented as

`\sigma _(\= x ) = (\sigma )/( √(n) )`

=>
`\sigma _(\= x ) = (1.93 )/( √(55) )`

=>
`\sigma _(\= x ) = 0.2602`

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

`P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 1 - P( (X - \mu )/(\sigma_(\= x )) > ( 2.10 - 2.45)/( 0.2602) )`

Generally
`(X - \mu )/(\sigma_(\= x )) = Z(The \ standardized \ value \ of \ X )`

`P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 1 - P( Z > -1.345 )`

From the z-table the value of

`P( Z > -1.345 ) = 0.91069`

So

`P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 1 - 0.91069`

`P(X \le 2.10 ) = 1 - P(X > 2.10 ) = 0.08931`

`P(X \le 2.10 ) = 0.08931`