Question

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

Answer 1

Part a

P(Z>k) = 0.3015

P(Z<k) = 1 - 0.3015

P(Z<k) = 0.6985

So, the value of k by using z-table is given as below:

k = 0.52

Answer: 0.52

Part b

P(k < Z < -0.18) = 0.4197

P(Z< -0.18) - P(Z<k) = 0.4197

0.4286 - P(Z<k) = 0.4197

(by using z-table)

P(Z<k) = 0.4286 - 0.4197

P(Z<k) = 0.0089

So, the value of k by using z-table is given as below:

k = -2.37

Answer: -2.37