SOLUTION: Assume that z-score are normally distributed with a mean of 0 and a standard deviation of 1 if p(-b<z<b)=0.2842, find b

Question 1163425: Assume that z-score are normally distributed with a mean of 0 and a standard deviation of 1 if p(-b
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import scipy.stats as stats
# Given P(-b < Z < b) = 0.2842
# 2 * P(Z < b) - 1 = 0.2842
# P(Z < b) = (1 + 0.2842) / 2
prob_symmetric = 0.2842
p_cumulative = (1 + prob_symmetric) / 2
b = stats.norm.ppf(p_cumulative)
print(f"{p_cumulative=}")
print(f"{b=}")
```
```text?code_stdout&code_event_index=2
p_cumulative=0.6421
b=0.3640776846923426
```
To find the value of $b$ for the standard normal distribution where $P(-b < Z < b) = 0.2842$, we follow these steps:
### 1. Understand the Symmetry
The standard normal distribution is symmetric around the mean (0). Therefore, the area in the center is given by:
$$P(-b < Z < b) = \Phi(b) - \Phi(-b)$$
Since $\Phi(-b) = 1 - \Phi(b)$, we can rewrite this as:
$$P(-b < Z < b) = \Phi(b) - (1 - \Phi(b)) = 2\Phi(b) - 1$$
### 2. Set Up the Equation
Substitute the given probability ($0.2842$) into the equation:
$$2\Phi(b) - 1 = 0.2842$$
$$2\Phi(b) = 1.2842$$
$$\Phi(b) = \frac{1.2842}{2}$$
$$\Phi(b) = 0.6421$$
### 3. Find the z-score ($b$)
We now look for the value $b$ such that the cumulative probability to its left is $0.6421$. Using a standard normal distribution table or a calculator:
$$b = \text{invNorm}(0.6421)$$
$$b \approx 0.3641$$
**Final Answer:**
The value of $b$ is approximately **$0.3641$**.
RELATED QUESTIONS
Assume that z-scores are normally distributed with a mean of 0 and a standard deviation... (answered by ewatrrr)

Assume that z-scores are normally distributed with a mean of 0 and a standard deviation... (answered by ewatrrr)

Assume that z-scores are normally distributed with a mean of 0 and a standard deviation... (answered by ikleyn)

assume that z-scores are normally distributed with a mean of 0 and a standard deviation... (answered by stanbon)

Assume that Z-scores are normally distributed with a mean of 0 and a standard deviation... (answered by ewatrrr)

Assume that z-scores are normally distributed with a mean of 0 and a standard deviation... (answered by stanbon)

Assume that z scores are normally distributed with a mean of 0 and a standard deviation... (answered by ewatrrr)