This question is incomplete, the complete question is;
One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 45 individuals who consider themselves to be avid Internet users results in a mean time of 1.98 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.98 hours or less from a population whose mean is presumed to be 2.35 hours.
Assume that the standard deviation for "time spent watching television on a weekday" is 1.93 hours.
Answer:
the likelihood of obtaining a sample mean of 1.98 hours or less is 0.0985
Step-by-step explanation:
Given the data in the question;
x" = 1.98
μ = 2.35
s = 1.93
n = 45
so, p( x" < 1.98 ) = P( (x" - μ)/(s/√n) )
we substitute in our values,
p( x" < 1.98 ) = P( (1.98 - 2.35) / (1.93/√45) )
= p( -0.37 / 0.2877
= P( - 1.29 )
from standard normal table,
P( - 1.29 ) = 0.0985
Therefore, the likelihood of obtaining a sample mean of 1.98 hours or less is 0.0985
