Question
Let \( Z \sim N(0,1) \) Without using the normalcdf(...) function, evaluate \( P(0 \leq Z \leq 1) \) 0.66 0.34 0.32 0.68 0.16
Ask by Franklin Gray. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( P(0 \leq Z \leq 1) \approx 0.34 \)
Solution
We start with the definition of the cumulative distribution function (CDF) for the standard normal distribution:
\[
P(Z \leq z) = \Phi(z)
\]
For the interval \( 0 \leq Z \leq 1 \):
\[
P(0 \leq Z \leq 1) = \Phi(1) - \Phi(0)
\]
Since for the standard normal distribution:
\[
\Phi(0) = 0.5
\]
and
\[
\Phi(1) \approx 0.8413
\]
we have:
\[
P(0 \leq Z \leq 1) \approx 0.8413 - 0.5 = 0.3413
\]
Thus, the closest answer is:
\[
\boxed{0.34}
\]
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution The Deep Dive
To evaluate \( P(0 \leq Z \leq 1) \) where \( Z \sim N(0,1) \), we first look for the area under the standard normal curve between 0 and 1. The total area under the curve is 1, and we know from statistical tables or the properties of the standard normal distribution that \( P(Z \leq 1) \approx 0.8413 \) and \( P(Z < 0) = 0.5 \). Thus, we can calculate: \[ P(0 \leq Z \leq 1) = P(Z \leq 1) - P(Z < 0) = 0.8413 - 0.5 = 0.3413 \] Therefore, upon rounding, the probability \( P(0 \leq Z \leq 1) \approx 0.34 \), which is an option provided in your list. So, the answer is **0.34**.
