Q.32 The variable z has a standard normal distribution.
If P(0 ≤ z ≤ 1) = 0.34, then P(z2 > 1) is equal to (up to two decimal places) ________
Introduction
The standard normal distribution is widely used in probability and statistics.
In this example, we determine the probability that the square of a standard normal variable is greater than 1 based on given information.
The problem states:
Given: The variable z has a standard normal distribution.
If: P(0 ≤ z ≤ 1) = 0.34
Find: P(z² > 1) correct to two decimal places.
Understanding the Condition
The condition z² > 1 means z lies outside the interval -1 to 1.
Therefore:
P(z² > 1) = P(z > 1) + P(z < -1)
Use of Symmetry
The standard normal distribution is symmetric around 0.
We are given:
- P(0 ≤ z ≤ 1) = 0.34
- P(z ≤ 0) = 0.50
So:
P(z ≤ 1) = 0.50 + 0.34 = 0.84
Thus:
P(z > 1) = 1 − 0.84 = 0.16
By symmetry:
P(z < −1) = 0.16
Final Calculation
P(z² > 1) = P(z > 1) + P(z < −1)
= 0.16 + 0.16
= 0.32
Final Answer
P(z² > 1) = 0.32 (correct to two decimal places)
Conclusion
Using symmetry and the provided probability over the interval 0 to 1,
we find that the probability outside the range -1 to 1 is 0.32.
Therefore, the probability that z² exceeds 1 is 0.32.
