Question

According to Nielsen Media Research, the average number of hours of TV viewing by adults (18 and over) per week in the United States is 36.07 hours. Suppose the standard deviation is 9.7 hours and a random sample of 43 adults is taken. Appendix A Statistical Tables a. What is the probability that the sample average is more than 37 hours? b. What is the probability that the sample average is less than 36.5 hours? c. What is the probability that the sample average is less than 27 hours? If the sample average actually is less than 40 hours, what would it mean in terms of the Nielsen Media Research figures? d. Suppose the population standard deviation is unknown. If 66% of all sample means are greater than 34 hours and the population mean is still 36.07 hours, what is the value of the population standard deviation? (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.) ro é o o

Answer 1

Criven that hours u= 36:07 = 9.7 n=43 hours. - (a) Let sample average = X-u is is X standard normel distributio Pl X 37 & Pl X-36.07 37-36.07 ) 143 793 = P(Z > •6287) = 5- Plocze 620 7 =.5-2357 = 2643 we see normal table. 6 - 9.7 V43 P(x< 36.5) P(x-3607 - 36.5 - 36.07 ) 9.7 F43 = Plze -103 X 93) - P(2<--2907) - P(Z <--2307) = P(Z > •2907) = 5 - PLOKZ2907) by symmetric = 5 - 1141 3859 h P(X (27) = P(8-36:07 4 27 - 30-07) 9.7 43 = P(ZK =.5 6.1315) = P lz > 6.1315) - PLOCZK 6/3/5) = 5 .5

PC x < 40) Epl X - 36:07 < 40-36.07) 9.7 9.7 143 143 POZ< 2.6567 = 996/ Probability of lees that 40 hawes is very high. It means that 99.61% chance that people watches tv les them 40 hours. @ PC X734) = .66 P( X-3607 34 - 36.07 ) = 66 Plz> - 2:07 xV23) = 66 Plz > -13.5739 ) = 66 5+ PC-13-5733 <3 <O) = 66 PC OZ Z < 13.5733 ) = .46 o 13.5739 - 42 T = 32.31 /