Question

mean = 64.1 inches. standard deciation = 3.8 inches

answer a,b,c,d

23. Let X be the actual height, in inches, of an adult female. X is approximately normally distributed with mean of 64.1 inches and standard deviation of 3.8 inches. a) Compute the probability that a randomly selected adult female is shorter than 60 inches. b) What proportion of adult females is between 60 and 66 inches tall? c) If 75% of women are shorter than Alice, what is Alice's height? d) If 40% of women are taller than Liz, what is Liz's height? e) Find and interpret the 80 percentile of this distribution.

if 75% of women are shorter than alice, what is her height?

if 40% if women are taller than liz, what s her height?

E) find and interperet the 80th percentile of this distribution

Answer 1

3.8 - 641 X ~ N Drob. X~ N(M = 84.1, 8= 3.8) as Reg. prob. & P (x260) = P(x- 4 6orul = P(Z < 60-64.1) - P (2 L -1.08 ) = CN fo-u -1002 -0.5-pl6.625 1.08) 40.5 0.5 - 0.3699 - adult inches 0.1401 Pusing Normal tahler femates between 60 tall b> of proportion ' and 66 E p (60 § x 566) DE i love you ebbe anche) =P (fail Şz¢ 562 ) = f(-1-08 Z Ş 0.5) - plos 2505) + Plocz< 1.08 ) Speed Oistes et 0.3599 forsing Normal ? 0.1915 + 0 Nomes Spiral = 0.5514 T J fohle ] =

let Alice's the a height be se 41 . -0.15 P (24) : . ; 0.5 *-4 / ५-५) : 6. as P ( L x-suil) : 0.75 ___20_0) + (0524 2- ६५.1 ) = 2-8 ___ 7.8 0.25 50. Ko + P100 2 < x•bur / 3.6 ____ 095 7- 0 8-64.1+0 3. 3.8 ) [ ° 2 267)____ • F- - 05 - 00. using Hommal to > ___. 2.68 7-64.) रह - (११.02 2 50/- .. ११. 04 __ .35 .19 -४ ४.- 1 89. 0 तो mPP 5 6५.। + (0.68) ह. 6.684 n chy Spiral

D LL .. I = d) o let a lia's hight here. p@xx) 407.5 0.40 P(x-u > x-1) = 0.40 3 pſ2 > 2= 64.11 ) 30:40 2-64.1 3 -8 x os-llosze x-64.1) 20.40 o.5 - 71-64.1 3.8 24 x-64. I 229 264) 3 .B 05 Tko.53 - 3-8 0.50 -0.40 » POL 2 X-64.1) = 0 yleentheid ) = 0.40 Liesung = p os 220.28) using Hormal table x-64.1 0.25 X = 64.1 + (318)(0-25 - 65.05 inches. Spiral

D LL .. I = d) o let a lia's hight here. p@xx) 407.5 0.40 P(x-u > x-1) = 0.40 3 pſ2 > 2= 64.11 ) 30:40 2-64.1 3 -8 x os-llosze x-64.1) 20.40 o.5 - 71-64.1 3.8 24 x-64. I 229 264) 3 .B 05 Tko.53 - 3-8 0.50 -0.40 » POL 2 X-64.1) = 0 yleentheid ) = 0.40 Liesung = p os 220.28) using Hormal table x-64.1 0.25 X = 64.1 + (318)(0-25 - 65.05 inches. Spiral