Question 1160789: the average mark of candidates in an aptitude test was 138.5 with a standard deviation of 10.6. Three scores extracted from the test are, 178, 122, 100. What is the average of the extracted scoresthat are extreme values?
Answer by CPhill(2303)   (Show Source):
You can put this solution on YOUR website! ```python
mu = 138.5
sigma = 10.6
scores = [178, 122, 100]
z_scores = {x: (x - mu) / sigma for x in scores}
for x, z in z_scores.items():
print(f"Score: {x}, z-score: {z:.4f}")
```
```text
Score: 178, z-score: 3.7264
Score: 122, z-score: -1.5566
Score: 100, z-score: -3.6321
```
To determine which scores are extreme values (outliers), we calculate the standard score ($z$-score) for each value using the formula:
$$z = \frac{x - \mu}{\sigma}$$
where:
* Population mean ($\mu$) = $138.5$
* Standard deviation ($\sigma$) = $10.6$
---
### Step 1: Calculate the $z$-scores
1. **For $x = 178$:**
$$z = \frac{178 - 138.5}{10.6} = \frac{39.5}{10.6} \approx +3.73$$
2. **For $x = 122$:**
$$z = \frac{122 - 138.5}{10.6} = \frac{-16.5}{10.6} \approx -1.56$$
3. **For $x = 100$:**
$$z = \frac{100 - 138.5}{10.6} = \frac{-38.5}{10.6} \approx -3.63$$
---
### Step 2: Identify the Extreme Values
In statistical analysis, an observation is conventionally considered an **extreme value (outlier)** if its $z$-score satisfies $\vert{}z\vert{} > 3$ (or $\vert{}z\vert{} > 2$ under broader outlier criteria):
* **$178$** has $z \approx +3.73$ ($> 3$), which is an **extreme value**.
* **$122$** has $z \approx -1.56$ (within $[-2, 2]$), which is a **normal value**.
* **$100$** has $z \approx -3.63$ ($< -3$), which is an **extreme value**.
The extracted scores that are extreme values are **$178$** and **$100$**.
---
### Step 3: Compute the Average of the Extreme Values
$$\text{Average} = \frac{178 + 100}{2} = \frac{278}{2} = \mathbf{139}$$
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