Answer)
We need to look into the standard normal z table to estimate the answer.
From z table, P(z>-0.52) = 0.7.
So, answer is -0.52
Find an approximate value of the standard normal random variable z, called zor such that P(Z...
Question 11 4 pts Find an approximate value of the standard normal random variable z, called to such that P (2>20) = 0.70. O -0.52 O -0.98 -0.47 -0.81 < Previous Next
(1 point) Find the value of the standard normal random variable z, called Zo such that: (a) P(Z <zo) = 0.8319 20 (b) PC-Zo <z<zo) = 0.5508 20 = (c) P(-20 <2<zo) = 0.748 zo = (d) P(z > Zo) = 0.2823 20 = (e) P(-20 <z<0) = 0.0283 Zo = (1) P(-1.5 <2<zo) = 0.7108 zo Note: You can earn partial credit on this problem.
(1 point) Find the following probabilities for the standard normal random variable z. (a) P(-0.81 <<0.42) (b) P(-1.14 <z < 0.5) (c) P(Z < 0.69) a (d) P(Z > -0.6)
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Find a value of a standard normal random variable Z (call it z0) such that (a) P(Z ?20) .30 (b) P(Z> z)16 (c) P(-20< Z<20).90
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.
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
Find the value of the standard normal random variable z, called z0 such that: (a) P(z≤z0)=0.7247 z0= (b) P(−z0≤z≤z0)=0.504 z0= (c) P(−z0≤z≤z0)=0.41 z0= (d) P(z≥z0)=0.0112 z0= (e) P(−z0≤z≤0)=0.1587 z0= (f) P(−1.21≤z≤z0)=0.6928 z0=
Find a value of the standard normal random variable z, call it zo, such that the following probabilities are satisfied. a. P(zszo)= 0.0185 b. P(-20 szszo) = 0.95 c. P(-20 szszo) = 0.99 d. P(-20 szszo)=0.8646 e. P(-20 Sz50) = 0.2501 f. P(-2<z<zo)= 0.9612 g. P(z>zo) = 0.5 h. P(z szo)= 0.0076
Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(1.22<Z<c)=0.0703 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places. 0 X $ ?