Find a value of the standard normal random variable z,call t z a. P(zszo)-0.0224
ID: 3064461 • Letter: F
Question
Find a value of the standard normal random variable z,call t z a. P(zszo)-0.0224 b. P(-20 52szo)-0.99 c. P-z0 5zszo)-0.90 such that the following probabilities are satisfied e. P-zo Szs0)-0.3447 f.Pl-2czczo)=0.9586 P(z Zo)-0.5 h. P(zszo)-0.0053 a. z0 (Round to two decimal places as needed) b.Zg.L(Round to two decimal places as needed.) c.4°L(Round to two decimal places as needed.) d. 20(Round to two decimal places as needed.) Round to two decimal places as needed) 0(Round to two decimal places as needed.) g.20(Round to two decimal places as needed.) h. 2(Round to two decimal places as needed)Explanation / Answer
Solution:
Find the value of the standard normal random variable z, call it z0, such that following probabilites are satisfied.
a. P (z z0) = 0.0224 = P(z z0) = 1-0.0224 = 0.9776 = P(z -z0) = 0.9776
From tables z0 = -2.01
b. P (-z0 z z0) = 0.99 = P (z z0)- P (z -z0) = P (z z0)- P (z z0) =
P (z z0)-(1- P (z z0))
P (z z0) = (0.99+1)/2 = 0.995
From tables z0 = 2.575
c. P (-z0 z z0) = 0.90
the procedure is the same that exercise b P (z z0) = (0.9+1)/2=0.95
From tables the nearest value is z0 = 1.64
d. P (-z0 z z0) = 0.8662
the procedure is the same that exercise b P (z z0) = (0.8662+1)/2= 0.9331
From tables z0 = 1.49
e. P (-z0 z 0) = 0.3447= P (z 0) - P (z -z0) = P (z 0) - P (z z0) =
P (z 0) - (1- P (z z0))
P (z z0) = 0.3447 + 1 - P (z 0)= 0.3447 + 1 - 0.5 = 0.8447
From tables z0 = 1.01
f. P (-2 z z0) = 0.9586 = P (z z0) - P (z -2) = P (z z0) - P (z 2) =
P (z z0) - (1- P (z 2))
P (z z0) = 0.9586 + 1 - P(z 2) = 0.9586 + 1 - 0.9773 = 0.9813
From tables the nearest value is z0 = 2.08
g. P (z z0) = 0.5 = 1- P (z z0)
P (z z0) = 0.5
From tables value z0 = 0
h. P (z z0) = 0.0053 = 1 – P(z z0)
P(z z0) = 1 - 0.0053 = 0.9947
From tables z0 = 2.56
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.