Question

Probability and I A recent Gallup Poll asked a simple random sample o cigarettes in the past week?" Suppose that in fact 20 question. The proportion of the sample who answer this, we simulated 1000 samples of size n=1600 from a populatie smoked cigarettes in the past week. The results of this simula the percentage column and write your answers as decimals done for you as an example.) om sample of 1600 American adults, "Have you, yourself smoked any ose that in fact 20% of all American adults would answer yes rican adults would answer"yes" if asked this "ple who answers "yes" will vary in repeated sampling. To investigate 1000 from a population in which 20% would answer yes they s of this simulation are provided in the table below. Please fill in swers as decimals rounded to three places. (The first two have been Frequency of the 1000 samples) 22 58 0.165 to 0.170 0.170 to 0.175 0.175 to 0.180 0.180 to 0.185 0.185 to 0.190 0.190 to 0.195 0.195 to 0.200 0.200 to 0.205 0.205 to 0.210 0.210 to 0.215 0.215 to 0.220 0.220 to 0.225 0.225 to 0.230 0.230 to 0.235 Percentage 0.001 0.006 0.022 0.058 0.089 0.146 0.178 89 146. 178 182 160 81 49 0.08 o oua 7 0.007 0.00 1.007 1,000

Answer 1

Solution : Thelooctuckion! - from Given information, n = 1600, P=0.20 mean = 0.00 and standard Esbor-/pt-P) ✓n - 0.20 (1-0.20) 1600 = 0.01 n normal Now, 68-95-991. rulee says that in Clove. 68% lies in the interval mean I 1. standad soody -0.19 to 0.21 95% lies in the interval meant standardera -0-18 to 0.22 996 lies in thie interval mean † 3. standard xror < 0.17 to 0.83 3 a) P( <0, 14) 20.16 0.17 0.18 0.19 0.2 0.2 0.22023

2) b. plộ 20.9) =0.001+0.006 + 0.022 +0.058+0.089 = 0.176 31a. P(0.18 21940.22) 30.95 b. plo.18 CÅ <0.22) -0.058 +0.089+ -- +0.049 = 0.943 0.17 0.18 0.2 021 0.22 0.23 4. No s. The result would have been same if the same size, n a AS 68-95-99-7 Rull takes into consideration the normal curve and the data fits normal Curve if its size is very large

3 The results of port a and 6 could have been same, as sample size is Kept as n=1600. 7. The law of arge numbers 1. Answer: option ® a. From Given information, mean = 0.20, standard deviation=0.01 To find PCÊ x0.22) i.c., to find plỘ >0.2048x0.01) Now 68-95-997. gule says that, in normal curve. 68% lies in the interval mean II standard Emor 0.19 to 0.21 959. lies in the interval meu I a standland Esra 0.18 to 0.22 qqe lies in the interval meant 3 standard Enor 0.17+ 0.23

PCÊ 20.22) =P (Ø20.2+280.01) : 0.025 0.025 * OY K-*.954 . 0.2-0.05 0.2-0.02 ** .0.2-0.017 0.2 toidl 0.2+0.02 to 0.2 +0:03 placentage 0.01 .6 0-06 0.osebni 0.058 bl from above table class for Å frequency 0.165 to 0.170 din 0.170 to 0.17.5 0.175 to 0.180 0.180 to o. 185 0.185 to 0-190 0.190 to 0.145 146 0.195 to 0.200 178 0.200 to 0.205 182 0.2o5 to 0.210 160 0.210 to 0.215 81 0.915 to 0.220 49 0.920 to 0.225 17 0.285 to 0.930 1 - 89 0.089 0.146 0.182 o 0.160 0.081 0.049 0.049 0.017 0.007

0.930 to 0.235 .4 0.004 P.(? 30.22) 20.017 to.007 to.o4 10.028