Question

This The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d). (a) Do you think the variable weekly time spent watching television would be normally distributed? If not, what shape would you expect the 1.525 variable to have? 1.525+ he com OA. The variable weekly time spent watching television is likely symmetric, but not normally distributed. OB. The variable weekly time spent watching television is likely normally distributed, . C. The variable weekly time spent watching television is likely skewed right, not normally distributed D. The variable weekly time spent watching television is likely skewed iðt, not normally distributed. O E. The variable weekly time spent watching television" is likely uniform, not normally distributed. () According to a certain survey, adults spend 2.25 hours per day watching television on a weekday. Assume that the standard deviation for time spent wiching television on a weekday" is 1.93 hours. If a random sample of 60 adults is obtained, describe the sampling distribution of We mon amount of time spent watching television on a weekday 1 pproximately normal with ; -2.25 and 0-0 (Round to uxdecimal places as voed) (e) Determine the probability that a random sample of 60 adults results in a mean time watching television on a weekday of between 2 and 3 it worried 0.566 hours The probability is (Round to four decimal places as needed.) Click to select your answert questions vey of 1006 adults, a poling

Answer 1

b) $\mu _{\bar x}$ = 2.25

$\sigma _{\bar x} = \frac{\sigma}{\sqrt n}$

$= \frac{1.93}{\sqrt {60}} = 0.249162$

$\bar x$ is approximately normal wii $\mu_{\bar x}$ = 2.25 and $\sigma_{\bar x}$ = 0.249162

c) P(2 < $\bar x$ < 3)

$= P\left(\frac{2 - \mu}{\sigma/\sqrt n} < \frac{\bar x - \mu}{\sigma/\sqrt n} < \frac{3 - \mu}{\sigma/\sqrt n}\right)$

$= P\left(\frac{2 - 2.25}{1.93/\sqrt {60}} < Z < \frac{3 -2.25}{1.93/\sqrt{60}}\right)$

= P(-1 < Z < 3.01)

= P(Z < 3.01) - P(Z < -1)

= 0.9987 - 0.1587

= 0.8400

d) P($\bar x$< 1.88)

$= P\left( \frac{\bar x - \mu}{\sigma/\sqrt n} \leq \frac{1.88 - \mu}{\sigma/\sqrt n}\right)$

$= P\left(Z \leq \frac{1.88-2.25}{1.93/\sqrt{55}}\right)$

= P(Z < -1.42)

= 0.0778

Expected value = 1000 * 0.0778 = 77.8 = 78

Option - C) If 1000 different random samples of size n = 55 individuals from a population whose mean is assumed to be 2.25 hours is obtained, we would expect a sample mean of 1.88 or less in about 78 of the samples.

No, the internet user doesn't watch less television.