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Given a standard normal distribution, find the value of h such that:i- P(Z < h) = 0.4840ii- P(Z > h) = 0.3707iii- P(h < Z < - 1.4) = 0.0428

User Jwalker
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Given a standard normal distribution, let's solve for the following:

• (i) P(Z < h) = 0.4840

To find the value of h here, apply the standardized Normal deviate using the NORMSINV table.

We have:

NORMSINV(0.4840) = -0.04

Hence, the value of h is -0.04

P(Z < 0.04) = 0.4840

• ii). P(Z > h) = 0.3707

We have:

P(Z < h ) = 1 - 0.3707

P(Z < h) = 0.6293

Now, apply the NORMSINV function:

NORMSINV(0.6293) = 0.33

The value of h here is 0.33

P(Z < 0.33) = 0.6293

• iii). P(h < Z < - 1.4) = 0.0428

Here, we have:

P(h < Z < - 1.4) = P(Z < -1.4) - P(Z < h) = 0.0428

Use the NORMSDIST function to find P(Z < -1.4):

NORMSDIST(-1.4) = 0.0808

Thus, we have:

P(Z < -1.4) - P(Z < h) = 0.0808 - P(Z < h) = 0.0428

P(Z < h) = 0.0808 - 0.0428

P(Z < h) = 0.038

Apply the NORMSINV function to find h:

NORMSINV(0.038) = -1.78

Thus, we have:

P(-1.78 < Z < -1.4) = 0.0428

h = -1.78

ANSWER:

i). -0.04

ii). 0.03

iii). -1.78

User Dagrooms
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