Solution:
The value of Zα/2 on a standard normal distribution
such that
P(-Zα/2 ≤ Z ≤ Zα/2) = 0.98
here Zα/2 = Z0.02/2 = Z0.01 = 2.33 (Standard normal distribution table)
=> P(−2.33 ≤ Z ≤ 2.33) = 0.98
Find the value of a/2 on a standard normal distribution such that P(-2a/2 SZ Sza/2) =...
We can now use the Standard Normal Distribution Table to find the probability P(-0.25 sz s 1). 0.05 0.06 0.07 0.08 0.09 -0.2 0.4013 0.3974 0.3936 0.3897 0.3859 0.00 0.01 0.02 0.03 0.04 Using these 1.0 0.8413 0.8438 0.8461 0.8485 0.8531 The table entry for z = -0.25 is 0.00 and the table entry for z = 1 is values to calculate the probability gives the following result. PC-0.25 sz s 1) P(Z < 1) - P(Z 5 -0.25) 10....
Find the value 2* that satisfies each of the following probabilities for a standard normal random variable Z. (Round your answers to two decimal places.) (a) P(Z sz*) - 0.0485 (b) PIZ sz*) = 0.9515 (c) P(-x* SZ SZ") - 0.903
Standard Normal distribution. With regards to a standard normal distribution complete the following: (a) Find P(Z > 0), the proportion of the standard normal distribution above the z-score of 0. (b) Find P(Z <-0.75), the proportion of the standard normal distribution below the Z-score of -0.75 (c) Find P(-1.15<z <2.04). (d) Find P(Z > -1.25). (e) Find the Z-score corresponding to Pso, the 90th percentile value.
Suppose that Z is the standard normal distribution. Find P(Z<-1.81). Suppose that Z is the standard normal distribution. Find P(Z>2). Suppose that Z is the standard normal distribution. Find P(-1.95<Z<1.07). Suppose that Z is the standard normal distribution. What value of Z represents the 20th percentile?
1. If Y has a normal distribution with mean 116 and standard deviation 16, find P(Y < 87). Enter your answer as a decimal, not a percent. For example, if the answer is 0.3161, enter 0.3161, NOT 31.61%. 2. It takes two days for a coffee shop to receive a shipment of milk from its supplier (this time is called "lead time" in operations management). During those three days, milk demand, Y, follows a normal distribution with a mean of...
Given a standard normal distribution, find the value of k such that (a) P(Z > k) = 0.3050; (b) P(Z <k) = 0.0367 (c) P(-0.96 <Z <k) = 0.7221 Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table (a) k= (Round to two decimal places as needed.) (b) k = (Round to two decimal places as needed.) (c) k= (Round to two decimal places as...
For a standard normal distribution, find: P(-2.43 < z < -1.87) For a standard normal distribution, find: P(-2.43 <z<-1.87) Submit License Question 3. Points possible: 1 This is attempt 1 of 3.
4. Given a standard normal distribution, find the value of k such that a. P(Z > k) = 0.3015 b. P(k < < < -0.18) = 0.4197
For a standard normal distribution, find the value of z such that: P(0<Z<z) = 0.4901 A. -1.65 B. +2.33 C. -3.33 D. +1.38
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.