Question 1168059: Individual test scores in a large statistics class are normally distributed with ( m= 84.2 )
and (s = 12.8 ) Let X represent the test score of a randomly selected student from this
class. Find the probability P( 80 < X < 86 ). Your solution must include a completely
labeled sketch of the distribution with the area of interest shaded.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to find the probability $P(80 < X < 86)$, including a completely labeled sketch of the normal distribution:
**1. Understand the Problem:**
We have a normally distributed random variable $X$ representing test scores with a mean ($\mu$) of 84.2 and a standard deviation ($\sigma$) of 12.8. We want to find the probability that a randomly selected student's score falls between 80 and 86.
**2. Convert the Scores to Z-Scores:**
To find the probability using the standard normal distribution table or calculator, we need to convert the given scores (80 and 86) into z-scores using the formula:
$z = \frac{X - \mu}{\sigma}$
* For $X = 80$:
$z_1 = \frac{80 - 84.2}{12.8} = \frac{-4.2}{12.8} \approx -0.3281$
* For $X = 86$:
$z_2 = \frac{86 - 84.2}{12.8} = \frac{1.8}{12.8} \approx 0.1406$
So, the probability $P(80 < X < 86)$ is equivalent to $P(-0.3281 < Z < 0.1406)$, where $Z$ is the standard normal random variable.
**3. Find the Probabilities Using the Standard Normal Distribution:**
We need to find the area under the standard normal curve between $z_1 = -0.3281$ and $z_2 = 0.1406$. We can do this by finding the cumulative probabilities $P(Z < 0.1406)$ and $P(Z < -0.3281)$ and then subtracting the smaller from the larger.
* $P(Z < 0.1406) \approx 0.5557$ (using a standard normal distribution table or calculator)
* $P(Z < -0.3281) \approx 0.3713$ (using a standard normal distribution table or calculator)
Now, subtract the probabilities:
$P(-0.3281 < Z < 0.1406) = P(Z < 0.1406) - P(Z < -0.3281) \approx 0.5557 - 0.3713 = 0.1844$
**4. Sketch of the Distribution:**
```
Normal Distribution
/ \
/ \
/ \
/ \
/ \
/ \
/ \
----------------------------------------------------
-3s -2s -1s μ +1s +2s +3s Z-scores
-3(12.8) -2(12.8) -1(12.8) 84.2 +1(12.8) +2(12.8) +3(12.8) Raw Scores
(-38.4) (-25.6) (-12.8) (12.8) (25.6) (38.4) Deviations from Mean
55.8 68.6 71.4 84.2 97.0 109.8 122.6 Approximate Raw Scores
| |
80 86
^ ^
| |
z=-0.33 z=0.14
<----------------------- Shaded Area ----------------------->
```
**Labeling the Sketch:**
* **Horizontal Axis:** Labeled as "Test Scores (X)" and also showing the corresponding "Z-scores" below.
* **Vertical Axis:** Represents the probability density (not explicitly labeled with values, but the curve shows the shape of the distribution).
* **Curve:** A bell-shaped curve representing the normal distribution.
* **Mean (μ):** A vertical line drawn at the mean score of 84.2, labeled with "μ = 84.2".
* **Standard Deviations (s):** Approximate locations of 1, 2, and 3 standard deviations away from the mean are indicated on the raw score axis.
* **Scores of Interest:** Vertical lines are drawn at the scores 80 and 86 on the horizontal axis.
* **Corresponding Z-scores:** The z-scores calculated for 80 (-0.33 approximately) and 86 (0.14 approximately) are indicated below the respective raw scores.
* **Shaded Area:** The area under the curve between the vertical lines at 80 and 86 (or their corresponding z-scores) is shaded. This shaded area represents the probability $P(80 < X < 86)$.
**Answer:**
The probability $P(80 < X < 86) \approx \boxed{0.1844}$.
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