3)using excel and Given that z is a standard normal random variable, what is the value for z0 if: a. P(z > z0) = 0.12 b. P(z < z0) = 0.2 c. P(z > z0) = 0.25 d. P(z < z0) = 0.3
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=
a.) Find z0 such that P(−z0 < z < z0) = 0.6. (Round your answer to two decimal places.) z0 = b.) What percentile does −z0 represent? (Round your answer to the nearest whole number.) c.) What percentile does z0 represent? (Round your answer to the nearest whole number.)
(a) Find a z0 such that P(z > z0) = 0.0287. (Round your answer to two decimal places.) z0 = (b) Find a z0 such that P(z < z0) = 0.9099. (Round your answer to two decimal places.) z0 =
For standadrd normal random variable Z, (i) given p(Z < z0) = 0.1056, find z0-score, (ii) Given p(-z0 < Z < -1) = 0.0531, find z0-score, (iii) Given p(Z < z0) = 0.05, find z0-score.
a) Find a z0 such that P(z > z0) = 0.0250. (Round your answer to two decimal places.) b) Find a z0 such that P(z < z0) = 0.8944. (Round your answer to two decimal places.)
Consider using a z test to test H0: p = 0.2. Determine the P-value in each of the following situations. (Round your answers to four decimal places.) (a) Ha: p > 0.2, z = 1.46 (b) Ha: p < 0.2, z = −2.77 (c) Ha: p ≠ 0.2, z = −2.77 (d) Ha: p < 0.2, z = 0.25
An infinite slab of charge of thickness 2z0 lies in thexy-plane between z=?z0 andz=+z0. The volume charge density ?(C/m3) is a constant.1-Use Gauss's law to find an expression for the electric field strength inside the slab (?z0?z?z0).Express your answer in terms of the variables ?,z, z0, and constant ?0.2-Find an expression for the electric field strength above the slab (z?z0).Express your answer in terms of the variables ?,z, z0, and constant ?0.3-Draw a graph of E from z=0 toz=3z0.
for the sample space {w,x,y,z}, p(x)=0.2, p(y)=0.15, p({w,y})=0.7, p({x,z})=0.3. Find p(w), p(z), and p({w,x,z}), using the properties of probability.
Find the value of the standard normal random variable z, called z0 such that:(a) P(z≤z0)=0.7909P(z≤z0)=0.7909z0=(b) P(−z0≤z≤z0)=0.6892P(−z0≤z≤z0)=0.6892z0=(c) P(−z0≤z≤z0)=0.98P(−z0≤z≤z0)=0.98z0=(d) P(z≥z0)=0.1258P(z≥z0)=0.1258z0=(e) P(−z0≤z≤0)=0.1431P(−z0≤z≤0)=0.1431z0=(f) P(−1.22≤z≤z0)=0.4883P(−1.22≤z≤z0)=0.4883z0=