If X is a normal random variable with parameters ? =10,?2 =36, compute p{X>5}. T

Question

If X is a normal random variable with parameters ? =10,?2 =36, compute p{X>5}. The solution is as shown in the screenshot but we usually calculate p{X>5} as 1-p{X<=5}. In that case z=-0.83. We usually say p(z(-0.83))=1-p(z(0.83)). but here it sais p(z(-0.83))=p(z(0.83)). I have 2 questions:

1-following statement is correct or not? I have p{X>5} I should calculate 1-p{X<=5} then I calculate z and probability and decut it from 1. But for p{X<5}, I just calculate Z and proability

2-why p(z(-0.83))=p(z(0.83)) in this case? when should I consider p(z(-0.83))=1-p(z(0.83)) and when I should consider p(z(-0.83))=p(z(0.83))

= P(Z >-0.83). On using the symmetry of normal distribution where ? is commutative of standard normal variants P(X>5) P(Z> 0-83) (0-83) On using a normal distribution table, we get P(x>5)07967

Explanation / Answer

There is a mistake in the second part of solution. It should have been P(Z<0.83).

statement 1: Yes. You can calculate it the way you said. That is, you can calculate the z-score and corresponding probability for P(X<5) and then deduct it from 1.

Because, 1-P(X<5) = 1-P(Z<-0.83) = 1-[1-P(Z>-0.83)] = P(Z>-0.83) ( this is the basic property of z-score) which bring you to step 1 of the given solution.

Statement 2: As I have told you, there is some printing error in the solution. They have meant that, P(Z>-0.83) = P(Z<0.83). You add '1-' when you change either one of negative or greater than. When you change both -ve to +ve and > to <, there is no need to add "1 -".

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