Question

can u please answer these questions as soon as possible?

Given that z is a standard normal random variable, find z for each situation (to 2 decimals). a. The area to the right of z is 0.03. b. The area to the right of z is 0.045 1.70 C. The area to the right of z is 0.05. 1.64 d. The area to the right of z is 0.1 Television viewing reached a new high when the global information and measurement company reported a mean daily viewing time of 8.35 hours per household. Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household. a. What is the probability that a household views television between 3 and 8 hours a day (to 4 decimals)? 0.4281 b. How many hours of television viewing must a top 4% of all television viewing households (to 2 decimals)? household have in order to be in the hours c. What is the probability that a household views television more than 3 hours a day (to 4 decimals)?

Answer 1

Civen The Tig af zis o 03 - 0-03 Aea foつ to e od 0-03 88.1 The 七 te riahd zis c-o45 anea P o 04S Aea fro z to oo 0-045 1-70 のThe P(z 0-05 Aea to 00 0-05 1-645 O area 0-1 te oo - 0-1 Area ズ-128 N

La TV vieco hnse hdld Given a N noith A-8. 35 and 2.5 prelabili bad a honsehold vieios to lenatoo beteen 3 and 8 hous P(3 sa58) The der a -8-35 - 8-3 <Z < 2.S 2 5 (-2.14 <z-0. 1 Area from -214 to -0.14 24 = Area tro -2.14 to - Area hom o-14 to -0.98 38- 0.5557 0.4281

Let a, The Iives iven PC no. of Cstetvieaorチ haus topoe =0.04 (だく)」 P(7 PZ7 ,8-3s 2 5 =004 z,) 647 Prea hon z, to ad 004 1.75 SE S10 175 2 5 2.5 X 1-7S ) +83512 725 12.73 lolevision viegny honss of A2-73 honse hold Wave n onder to be in the lop 4/ of all takvisieo Vieuor honse holds must NO

The prebabiliy tbal Vieos teleis'on 3 honss a dex P/273) 3-8 - P (7 - P(z-2.2) Area rom-2-2 to a 2.5 -2.2 O - 0.9861