The amount of time adults spend watching television is closely monitored by firm

Question

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) The variable "weekly time spent watching television" is likely skewed left, not normally distributed. The variable weekly time spent watching television" is likely skewed right, not normally distributed. The variable "weekly time spent watching television" is likely normally distributed. 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 watching television on a weekday" is 1. 93 hours a random sample of 50 adults is obtained, describe the sampling distribution of x^OverBar, the mean amount of time spent watching television on a weekday. x^OverBar is approximately normal with mu_x = 225 and sigma_x = 0.272943 (Round to six decimal places as needed) Determine the probability that a random sample of 50 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is 08172 (Round to four decimal places as needed.) 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 moan time of 1.86 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.86 hours or less from a population whose mean is presumed to be 2.25 hours. The likehood is _____. (Round to four decimal places as needed.)

Explanation / Answer

a)option C is correct

b) mean =2.25

and standard error of mean =standard deviation/(sample size)1/2 =0.272943

c) P(1<X<3) =P((1-2.25)/0.272943<Z<(3-2.25)/0.272943)=P(-0.9159<Z<2.7478)=0.9970-0.1798 =0.8172

d)for 45 sample size standard error =0.287707

hence P(X<1.86)=P(Z<-1.3555)=0.0876

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