Question

2) Women's heights are normally distributed with a mean of 64.1 in, and a standard deviation of 2.5in. a) What percentage of adult women can fit through the doors on the Mark VI monorail (find the height of the doors on the Monorail in the chapter 5 notes)? b) Does the answer to part a mean that all women are under 6 ft tall? If not, explain the probability in a complete sentence by converting it to a fraction. c) The tallest 10% of women are all over inches. inches tall which is

Answer 1

X: women's height ~ N(H = 64., 6 = 2.5) mark v monorail has height 72 in P( women caufit) - P(X <72) = pl zk 72 -64.1) 12.5 Salzk3.14) = 0.9994 No aus to past (a) does not mean that are women are undu oft tall PCXC72Cja) - 9992 Topo out of 1000 women, we expect 9992 to be under oft & & to above 6ft. Tallest 10% of women are over to in say >> P(x>20) = 0.16 => P(1272-4) +0.10 3p(271.2810) using z table zole " P(Z > 20-4)= 0.10 =) | Tw-u 1.2816 | = | a 07 64.1+ 1.2816x2.5 t 67.30 in and 7.3 in lie 5ft