Question

For a standard normal distribution, find: P(-1.64 < z < 0.2)

Answer 1

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; brainly.com/question/4621112

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