Question

(a) What is the probability that the size of a single droplet is less than 1410 um? At least 900 um? (Round your answers to four decimal places.) less than 1410 um at least 900 pm (b) What is the probability that the size of a single droplet is between 900 and 1410 pm? (Round your answer to four decimal places.) (C) How would you characterize the smallest 2% of all droplets? (Round your answer to two decimal places.) The smallest 2% of droplets are those smaller than um in size. (d) If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1410 um? (Round your answer to four decimal places.) 5:39 PM

Answer 1

Answer:
a) PC x< 1410) Norewol des trobeeteon el = 1050 o 150 We convert thes to using standared normal - lll گZ M10-1050 150 5 2:40 0.008198 P(Z 22.40) P(x 2 14101) PL * =900) 20.008198 o = 150 Norayau destolbution rel=1050 since, these le contenuove dee tri bie teory, P( X = good We convert this SPC *> 900) to e tandard normal eingz X-ll ZF 900-1050 ngo I-1 p (x >900)=P(Z >-1) sl-PLZ-1)

-1-0.8913 0.1587 PC x 2900) = 0.1587 b) P(900 LX21410) Nonyar des droe betoom ll. 5. 1050 o=150 We converet thals to standard gooyou wong Zs X-ll zit 900-1050 150 - 1410-1050 22 S 150 = 2:40 P(-1 {2<2.40) Area between - 1 and 2.40 P(x & Xe )^4 + . 2 ) s 2 100 s0.02 l=1050 150

3 X-ll E 0.02 orel) 3 Plzq x xo-ll. :) - = 0.02 s Xo- En v Norm 10:02) xosllt 0x en Norry (0:02) Xo = 10.50 +04-2.054) ñ 741.9 s. The smallest 2o. of doroplets are those smaller than 7.4205 lum en fize. d) The probability of one droplet exceeding 191044 84 M = 1050 o = 150 PCX >14101- plz>1410-1050) 150 P. (242.40) 0.0082 so.0082 use the benomear probability with oqship: P(X>1). = 1- P(x=0 p(x-7)= 500 (0.008 270(1-01008 2755 10 10.9997) 0.9597 X1 s 0.9597 I-p (x=0)=100095971-bc0yog Ang