Question

1. (5 points) Suppose Z is a random variable that follows the standard normal distribution. a) Find P(Z > 0.45). b) Find P(0.7 SZ 1.6). c) Find 20.09. d) Find the Z-score for having area 0.18 to its left under the standard normal curve. e) Find the value of z such that P(-2SZS2) -0.5.

Answer 1

Colation in (subpose znis random Kaviable that follocons : Pescuired information iągilan by @ Find pt 7200.15] in inte hakes to obtained the probability by aising given dat is Eler >probablite ks Hz? 0.418] -) 1- Plzco.c13] >(-0.6766] 0.3264 Therefore, is 0.8264) Rezcuired probability Dfind pto.7s7c 16]:- nk have to obtained the probability that is Gere, > probabi Cileg is PEO.ists116] pt t<l.6– pfz<0.1] =>6.2452] – 60.4580]

6. 1679 Therefore, Pesuired probability is 6.1842 © Find too.09) Valca is, lub have to obtained the given Here → ZX=Zt0.09] T. 6408 Therefore, Zco.oq) = T1.6406 area o.16 to left under i @ Find the 7-score for having 0.16 to lale have to obtained the z Glore for halling area lett under area is, Here, plz<2] = 0.16 =) 2=760.18] z=0 0. 91501

3 Therefore, Z [core. is F0.91541 (2) © Find the Kake of pezszcz]=0.5" ale have to obtained "₂" value by using given data its, Efere z-Value is pt-Zs7st] =0.5 -p+z<23– ptz--23503 >ptz<23-ti-ptz<z]] =0.5 -) PtZ<71- 1tptz<23=0.5 =) o Plz=z] -120.5 sopEZ<7)= ht0.5 => 0 ptzez] =1.5 -> ptZZJ-[-50ls

+ -> PEZ<z)=0 45 > - 10:50 Co.67451 Therefore, 0.6745 Z value is