Question

Question 1 (15 marks)

According to Nielsen Media Research, the average number of hours of TV viewing per household

per week in the United States is 50.4 hours.

(a) Suppose the population standard deviation is 11.8 hours and a random sample of 42 U.S.

household is taken, what is the probability that the sample mean TV viewing time is between

47.5 and 52 hours?

(b) Suppose the population mean and sample size is still 50.4 hours and 42, respectively, but

the population standard deviation is unknown. If 72% of all sample means are greater than

49 hours, what is the value of the unknown population standard deviation?

(c) What is the result of part (a) if the sample only consists of 5 households? Explain.

Answer 1

OLURU Let X denotes exrerage number huu of houn of C TV te Viewing per household, x has mean 50.4 @ S.D: 11.8, M: 42 Probability that the sample mean of TV viewing time is between 47.5 g 52 ei P (47.5<7 <5e) TNN (4: 50.4 ó- ) (41.5-50.4 xall 52-50.4. ::p ( 47.5 <*.<52) =P ( Mec livielas =P /-1.59272 <2< 0.8787) Les :p (-1.59276260) + P COC2C0.8787 -1.59 270 0.3147 : 0.444) + 0.3078 = 0.1519

LUULA 6 l = 50.9, n=42 À NN (4,250.4, P ( ² >99) = 0.72 - 50.4 49 - 50.41 1 0.70 1842 / - 49-50.9. p(28 415 0.70 0.70 ) from normal table, the ardinate corresponding to probability -0.53 0.70 ::P (2) -0.53 ) =0.70 ::. 49 - 50.4 6/542 = -0.53 -1.4 Omãe : -0.53 1.4 - Co 1.4 x 742 0.53 Os 17.1189 .; SD = 17.1189

Result of part @ if n=5 l=go. 4, 6= 11.8, n=5 . N N (50.4, 11.8%) P (47.5 < t < 52) e as follow. 747 - 50.4 X - 50.4 ( 71.875 9516 < 71.6115) 52-50.41 11.8/15 11.8/55 : P ( -0.549 5 < 2 < 0.3 032 ) : P (-0.54952220) + P (O<Z (0.3034 E 0. 2088 + 0.1179 -0.5495 0 0.3032 0.3267 P ( 47.5 < F< 52) = 0.3267
From a and b it is seen that if we reduce sample size then probability is decreases.