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Sagot :
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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