Question

During the flu season, flu patients can be contagious for 3 - 5 days and on average spread it to 6 people (SD = 3.4). o Patient A used proper preventative measures as soon as he started showing symptoms. He spread it to 2 people. What % of flu patients spread the virus more than A did? O Patient B did the opposite and became a "super spreader". B was in the top 3% in terms of how many people he had spread the virus to. Approximately how many people did B infect?

Answer 1

Here mean (M) = 6 and standard deviation co= 3.4 (کد )Nه لا i.l. XN NCG,3.4) a) patient A spread i for 7=2" flu to 2 people. 2 = x - M 2:-1,18 M6 20 2-6 3.4 = -1.1764 Z= -1.18 Now, using z table, the area under the curve to the Left oto Z=-1,18 is 0.1190 Total area under the normal cuore . Area of shaded region = 1-0.1190 = 0.881 .. Area of shaded region = 0.881 in 88.1 % of flu patients spread the virus more than A. .اون b) ß was in ton 3). The shaded area in the curve is 31. area. Therefore remaining area is 97% 0 - Now we have to - find value of ne. /

Now, using a table, we can find the I score corresponding to 941, that is o.aroo Now in a table value of are a 0.9700 lie in between 0.9699 and 0.9706 The 2 scores corresponding to these values are 1.88 and 1.89 respectively: :: The required a score is average of 1.88 and 1.89 . The required a score = 1.885 Now, z = 2-M x=20 tu = 1.885 (3.4) + 6 = 120409 i na 12 Therefore we can say that B infected appronimately 12 people.