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How do you find the number z such that 93 percent of all observations from a standard Normal distribution are greater than z?

The tables for Z-scores are given in the form of P = Prob(Z < z) for various value of P and z. Since Prob(Z > z) = 0.93 > 0.5, then by symmetry, z < 0. So suppose z = -a where a > 0 Now Prob(Z > -a) = 0.93 is the same as Prob(Z < a) = 0.93 [because the standard Normal is symmetric]. therefore, from the tables, a = 1.4758 (approx) and so z = -1.4758 (approx).


What is the value of z if the area between -z and z is 0.777?

To find the value of ( z ) such that the area between (-z) and (z) under the standard normal distribution curve is 0.777, we need to determine the corresponding cumulative probability. The area between (-z) and (z) represents the central portion of the distribution, so we can express this as: ( P(-z < X < z) = 0.777 ). This means that ( P(X < z) - P(X < -z) = 0.777 ). Since the distribution is symmetric, ( P(X < -z) = 1 - P(X < z) ), leading to ( 2P(X < z) - 1 = 0.777 ) or ( P(X < z) = 0.8885 ). Using a standard normal distribution table or calculator, we find that ( z \approx 1.175 ).


Z plus 6 equals -5?

1


What is the absolute value of the difference of z and 8?

|z - 8| or whichever of (z - 8) and (8 - z) that is non-negative.


How do you compute the probability pY and gtu given a random variable Y has a normal distribution with mean m and variance s2.?

You find z = (u - m)/s and the look up the Pr(Z > z) in the tables of cumulative probability of the normal distribution. Note that the tables only give the probabilities for Pr(Z <= z) for z >=0. So Pr(Z > z) = 1 - Pr(Z<=z). Also, if z < 0 then you will need to use the symmetry of Z about the value 0.