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Sagot :
For a standard normal distribution, the probability of the 2 - scores P(-1.64 < z < 0.2) is 0.52876
How to find the p-value from 2 z-scores?
We want to find the p-value between 2 z-scores expressed as;
P(-1.64 < z < 0.2)
To solve this, we will solve it as;
P(-1.64 < z < 0.2) = 1 - [P(z < -1.64) + P(z > 0.2)]
From normal distribution table, we have that;
P(x < -1.64) = 0.050503
P(x > 0.2) = 0.42074
Thus;
P(-1.64 < z < 0.2) = 1 - (0.050503 + 0.42074)
P(-1.64 < z < 0.2) = 0.52876
Thus, For a standard normal distribution, the probability of the 2 - scores P(-1.64 < z < 0.2) is 0.52876
Read more about p-value from z-score at; https://brainly.com/question/4621112
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