Let A and B be constants such that A < B. Let z be a standard normal
random variable. For what values of A does P(A<z<B)approximately
equal P(z<B)?
Pick ALL that apply.
a. A = -9
b. A = -26
c. A = 0
d. A = 9
e. None of these are correct.
This is how I'm reasoning this question:
B is not specified, but B > A.
a) A=-9, let B=-8.9
So, that's why I think the answer would be A and B. But I have a feeling I
am wrong.
To find the values of A for which P(A < z < B) approximately equals P(z < B), we need to consider the properties of the standard normal distribution.
In the standard normal distribution, P(z < B) represents the probability of z being less than a certain value B. This probability can be found using a z-table or a statistical software.
We can rewrite P(A < z < B) as P(z > A) - P(z > B) or as 1 - P(z < A) - P(z > B), because the standard normal distribution is symmetric around 0.
We want P(A < z < B) to approximately equal P(z < B), meaning that P(z > A) - P(z > B) should be close to 1 - P(z < A) - P(z > B).
Simplifying, we get P(z > A) ≈ 1 - P(z < A).
Now, considering the given options:
a) A = -9: P(z > -9) ≈ 1 - P(z < -9)
b) A = -26: P(z > -26) ≈ 1 - P(z < -26)
c) A = 0: P(z > 0) ≈ 1 - P(z < 0)
d) A = 9: P(z > 9) ≈ 1 - P(z < 9)
Since the standard normal distribution is symmetric around 0, we know that P(z > A) and P(z < A) have the same value. Therefore, in all the given options, P(z > A) ≈ 1 - P(z < A).
Hence, the answer is e) None of these are correct, as none of the options satisfy the condition P(A < z < B) approximately equaling P(z < B).