A)Find the following probabilty for a standard normal randomvariable z: P(-1.96

Question

A)Find the following probabilty for a standard normal randomvariable z: P(-1.96 z < -0.33)
For the value -1.96 I got found it to be 0.0250 and for -0.33I got 0.3707
Then to find the difference between the z scores I subtracted0.3707-0.0250 = 0.3460 was my answer
B) Find a zo such that P(z > zo) = 0.9750 I'm not sure how to do part B
P(-1.96 z < -0.33)
For the value -1.96 I got found it to be 0.0250 and for -0.33I got 0.3707
Then to find the difference between the z scores I subtracted0.3707-0.0250 = 0.3460 was my answer
B) Find a zo such that P(z > zo) = 0.9750 I'm not sure how to do part B

Explanation / Answer

Part B:
We are given a standard normal distribution. We wish to find Az0 such that P(z > z0) = .9750, which is equal to saying 1 - P(z < z0) = .9750 So P(z < z0) = .025 Interestingly enough, you used this value in part (a). Thevalue of Z(-1.96) = .025, so your answer is z0 = -1.96
If you did not already know this value, you would solve for 1- Z(-z0) = .025, since all the values in the table for a standardnormal distribution are .5 or above.

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