In: Statistics and Probability
A.C Neilson reported that children between the ages of 2 and 5 watch an average of 25 hours of television per week. Assume that the variable is normally distributed and the standard deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomly selected, find the probability that the mean of the numbers of hours they watch television will be greater than 26.3 hours.
Solution:
Given that ,
= 25
= 3
A sample of size n = 20 is taken from this population.
Let
be the mean of sample.
The sampling distribution of the
is approximately normal with
Mean
=
= 25
SD
=
= 3/
20
= 0.67082039325
find the probability that the mean of the numbers of hours they watch television will be greater than 26.3 hours.
i.e. Find P(
> 26.3)
P(
> 26.3)
= P[(
-
)/
> (26.3 -
)/
]
= P[ (
-
)/
> (26.3 - 25)/ 0.67082039325]
= P[Z > 1.9379255805]
= P[Z > 1.94]
= 1 - P[Z < 1.94]
= 1 - 0.9738 ( use z table)
= 0.0262
P(
> 26.3) = 0.0262