Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of so that the following is true.
P(-0.78≤Z≤c)=0.7657

Respuesta :

For a standard normal distribution, the value of the z-score represented as c in P(-0.78≤Z≤c)=0.7657 is 0.0166

How to find the p-value from 2 z-scores?

We want to find the p-value between 2 z-scores expressed as;

P(-0.78 < z < c) = 0.7657

To solve this, we will solve it as;

0/7657= 1 - [P(z < -0.78)  + P(z > c)]

From normal distribution table, we have that;

P(z < -0.78) =  0.217695

Thus;

0.7657 = 1 - (0.217695 + c)

c = 1 - 0.7657 - 0.217695

c = 0.0166

Thus, For a standard normal distribution, the value of the z-score represented as c in P(-0.78≤Z≤c)=0.7657 is 0.0166

Read more about p-value from z-score at; brainly.com/question/4621112

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