The probability is 0.8577 . (Round to four decimal places as needed.)
(d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 35 individuals who consider themselves
to be avid Internet users results in a mean time of 1.91 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.91 hours or less from a
population whose mean is presumed to be 2.35 hours.
The likelihood is . (Round to four decimal places as needed)
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UpStudy AI · Accepted Answer

To determine the likelihood of obtaining a sample mean of 1.91 hours or less from a population whose mean is presumed to be 2.35 hours, we can use the Z-score formula for a sample mean. The Z-score is calculated as follows:

$$
Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}
$$

Where:
- $$\bar{X}$$ = sample mean (1.91 hours)
- $$\mu$$ = population mean (2.35 hours)
- $$\sigma$$ = population standard deviation (not provided, so we will assume it is known or we will need to estimate it)
- $$n$$ = sample size (35)

Since we do not have the population standard deviation ($$\sigma$$), we cannot calculate the Z-score directly. However, if we assume a standard deviation, we can proceed with the calculation.

Let's assume a hypothetical standard deviation. For example, if we assume $$\sigma = 1.0$$ hours, we can calculate the Z-score as follows:

1. Calculate the standard error (SE):
$$
SE = \frac{\sigma}{\sqrt{n}} = \frac{1.0}{\sqrt{35}} \approx 0.1690
$$

2. Calculate the Z-score:
$$
Z = \frac{1.91 - 2.35}{0.1690} \approx \frac{-0.44}{0.1690} \approx -2.605
$$

3. Now, we can find the probability associated with this Z-score using the standard normal distribution table or a calculator.

Using a Z-table or calculator, we find:
$$
P(Z < -2.605) \approx 0.0045
$$

Thus, the likelihood of obtaining a sample mean of 1.91 hours or less from a population whose mean is presumed to be 2.35 hours is approximately **0.0045** (rounded to four decimal places).

If you have a specific standard deviation value, please provide it, and I can recalculate the likelihood accordingly.
The likelihood of obtaining a sample mean of 1.91 hours or less from a population with a mean of 2.35 hours is approximately 0.0045.

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