Question

The amount of time adults spend watching television is closely monitored by firms becayse this helps to determine advertising pricing for commercials. Compete parts (a) through (d).
a) Do you think the variable "weekly time spent watching television" would be normally distributed? Yes or No.
If not, what shape would you expect the variable to have? Skewed Left, Skewed Right, Uniform or Symmetric?
b) According to a certain survey, adults spend 2.35 hours per day watching television on a weekday. Assume that the standard deviation for "time spent watching TV on a weekday" is 1.93 hours. If a random sample of 40 adults is obtained, describe the sampling distribution of x-bar, the mean amount of time spent watching TV on a weekday.
Mean =
A) 2.35
B) 1.89
C) 2.25
SD = (round to six decimal places as needed)
c) Determine the probability that a random sample of 40 adults results in a mean time watching television on a weekday of between 2 and 3 hours.
The probability is ____ .
d) One consequence of the popularity of the Internet is that it is thought to reduce TV watching. Suppose that a random sample of 35 people who consider themselves avid Internet users results in a mean time of 1.89 hours watching TV on a weekday. Determine the likelihood of obtaining a sample mean of 1.89 hours or less from a population who mean is presumed to be 2.35 hours.
The likelihood is ____ .
Based on the result obtained, do Internet users watch less TV? Yes or No.

Answer 1

Answer:

Yes based on the result 0.4841 the internet users watch less TV because the mean would be placed at 0.5 and it is less than 0.5

Step-by-step explanation:

a)The variable "weekly time spent watching television" is normally distributed and is skewed right.

b) Mean = x` 2.35 hours

Standard deviation = s/√n = 1.93/√40=1.93/6.3245553= 0.3051598

c) P(2<X<3) = P (2-2.35/ 0.3051598< Z< 3 -2.35/ 0.3051598)

                    = P ( -1.14694 < Z <2.13003)

                     = 0.3729 + 0.4830

=0.8559

So the probability is 0.8559

(0.8559 we check the value of 2.13 from the normal distribution tables and add with the value of 1.14 to get the in between value -1.14694 < Z <2.13003)

d) Here n = 35 , s= 1.93 , mean = 2.35 and x= 1.89  

So Putting the values

P (X ≤ 1.89) = P (Z ≤ 1.89- 2.35/ 1.93 / √35)

              = P ( Z ≤ -0.238341/ 5.9160797)

                  = P ( Z ≤ -0.04028)

= 0.5 - 0.0159

= 0.4841

Similarly again subtracting from 0.5 the value from normal distribution table to get less than or equal to value.

Yes based on the result 0.4841 the internet users watch less TV because the mean would be placed at 0.5 and it is less than 0.5