Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
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
#SPJ1
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
