Solution:
b)
P( -1.69 Z z0 ) = 0.9545
P( Z z0 ) = 0.9545 + P( Z -1.69 )
P( Z z0 ) = 0.9545 + 0.025
P( Z 2.044 ) =0.9795
z0 = 2.044
c)
P( Z < z0 ) = 0.9952
P( Z < 2.59 ) = 0.9952
z0 = 2.59
d)
P( Z < z0 ) = 0.8413
P( Z < 1) = 0.8413
z0 = 1
e)
P( Z < z0 ) = 0.0287
z0 = -1.9
f)
P( Z < z0 ) = 0.6179
z0 = -0.87
and
z0 = 0.87
b. P(-1.69 SZ SZ) = 0.9545 C. P(Z > 20) = 0.0048 d. P(Z, SZS 2.00)...
Please show all work for each part. Thank you!! ECO 578 Summer 11, 2019 Page 2 of 12 2. Given the following probabilities, find Z, and please draw the shading the area Show your work PZSZ) = 0.2119 Please draw graphs b. P(-1.69 SZ SZ) = 0.9545 c. P(Z > Z) = 0.0048 d. P(Z SZ S 2.00) = 0.1359 c. P(Z > Z) = 0.9713 f. P(-2, SZ SZ) = 0.2358
SHOW ALL WORK PLEASE 2, Given the following probabilities, find Zo and please draw the shading the area Please draw graphs Show your work P(Z Zo) 0.2119 a. P(-1.69 ZZ) = 0.9545 b. P(Z> Z) 0.0048 C. - d. P(Zo Z S 2.00) = 0.1359 P(Z> Zo) 0.9713 e. P(-ZoS 0.2358 f.
7. Hint for c and d: given P(X S x) a percentage, we have P(Z Sz)the percentage. Then find the corresponding value for Z, and use the Inverse Transformation Let X be normally distributed with mean 120 and standard deviation σ 20. a. Find P(X3 86). b. Find P(80 <X3100). c. Find x such that P(Xx) 0.40. d. Find x such that P(X> x) 0.90.
In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.) Ф(c)-0.9838 (a) (b) P(O SZ sc) 0.2967 (c) PC s Z)0.1230 (d) P(-c c)-0.6424 Z (e) P(c s IZ1)-0.0160 In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.) Ф(c)-0.9838 (a) (b) P(O SZ sc) 0.2967 (c) PC s Z)0.1230 (d) P(-c c)-0.6424 Z...
-140 points DevoreStat9 4.E.028 Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate (Round your answers to four decimal places ( POS 2 5 2.79) (0) P-260 SZ50) (d) P(-2.50 SZ 52.60) (e) PZ 51.98) P(-1.5552) (o) RE-1.605 ZS 2.00) (W) 1.98 325 2.50) P103 O phere to search (b) POSZS2) (c) P(-2.60 SZ 30) (d) P(-2.60 SZS 2.60) (e) P(Z S 1.98) (f) P(-1.55 SZ) (9) P(-1.60 S ZS 2.00) (h)...
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.) (a) P(O SZS2.58) (b) PO Szs 1 (c)P-2.10 Zs o) (d) バー2.10 i 2.10) (e) P(Z 1.32) (f) P(-1.55 s Z) 9) P-1.10 s Z s 2.00) (h) P(1.32S Zs 2.50) O) P1.10 sz) (j) P(İZİS 2.50)
(4) Given Z N(0, 1) find the following: (a) P(Z 2 1.4) (b) P(Z> 0.75) (c) P(IZI S 2) (d) P(IZ 2 2) (e) Find z such that P(Z < z) = 0.11 (f) Find z such that P(Z > z) = 0.02
8) Suppose that p(z sz) = .0580 Using the z-table determine the closest value for zc a) -0.85 b) -1.25 c) -2.43 d) -1.57
Find a value of the standard normal random variable z, call it 20, such that the following probabilities are satisfied. a. P(zszo)=0.0502 b. P(-2o Szszo)=0.99 c. P(-zo szszo)=0.90 d. P(- zo szszo) = 0.8062 e. P(-Zo Szs 0) = 0.2593 f. P(-3<z<zo)=0.9654 g. Plz>20) = 0.5 h. Plz szo) = 0.0088
2. Calculate the probability from the z-table. a. P(Z > -0.45) b. P(Z < 2.10) c. P(0 < Z < 2.10) d. P(-0.45 < 3 < 2.10)