(d) 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 1.94 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.94 hours or less from a population whose mean is presumed to be 2.35 hours. The likelihood is $\square$. (Round to four decimal places as needed.)

Step 1 ：We are given a random sample of 45 individuals who consider themselves to be avid Internet users. The mean time they spend watching television on a weekday is 1.94 hours. We are asked to determine the likelihood of obtaining a sample mean of 1.94 hours or less from a population whose mean is presumed to be 2.35 hours.

Step 2 ：To solve this, we first need to calculate the z-score. The z-score is a measure of how many standard deviations an element is from the mean. We can calculate it using the formula \( z = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{population mean}} / \sqrt{{\text{{sample size}}}}}} \).

Step 3 ：Substituting the given values into the formula, we get \( z = \frac{{1.94 - 2.35}}{{2.35 / \sqrt{45}}} \).

Step 4 ：The calculated z-score is approximately -1.17.

Step 5 ：Next, we use the cumulative distribution function (CDF) of the normal distribution to find the probability associated with this z-score. This probability represents the likelihood of obtaining a sample mean of 1.94 hours or less.

Step 6 ：The corresponding probability is approximately 0.121.

Step 7 ：So, the likelihood of obtaining a sample mean of 1.94 hours or less from a population whose mean is presumed to be 2.35 hours is \(\boxed{0.1210}\).