Question 1185911: Question: Use Table A4 to compute the probability for any Normal random variable to take a value within 1.5 interquartile ranges from population quartiles.
P.S: Table A4 is a Standard Normal distribution Table.
I don't know how to approach this question because I didn't understand how to take a random variable within 1.5 IQR from population quartiles.
I would appreciate a more elaborate explanation and maybe a solution.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down this problem step by step.
**Understanding the Interquartile Range (IQR)**
The interquartile range (IQR) is a measure of statistical dispersion, or how spread out the data is. It's the range between the 25th and 75th percentiles of the data.
* **Q1 (First Quartile):** The value that separates the lowest 25% of the data.
* **Q3 (Third Quartile):** The value that separates the highest 25% of the data.
* **IQR = Q3 - Q1**
**Relating IQR to the Standard Normal Distribution**
When dealing with a standard normal distribution (mean = 0, standard deviation = 1), we can use a Z-table (like Table A4) to find the Z-scores corresponding to Q1 and Q3.
* The Z-score for Q1 is approximately -0.674
* The Z-score for Q3 is approximately 0.674
This means that the IQR of a standard normal distribution is 0.674 - (-0.674) = 1.348.
**Calculating the Probability**
1. **Determine the range:** We want to find the probability of a random variable falling within 1.5 IQRs from the population quartiles. Since the IQR is centered around the median (which is also the mean in a normal distribution), we need to calculate the range as follows:
* Lower bound = Q1 - 1.5 * IQR = -0.674 - 1.5 * 1.348 = -2.696
* Upper bound = Q3 + 1.5 * IQR = 0.674 + 1.5 * 1.348 = 2.696
2. **Use the Z-table:** Look up the cumulative probabilities for the upper and lower bounds in Table A4.
* P(Z < 2.696) ≈ 0.9965
* P(Z < -2.696) ≈ 0.0035
3. **Calculate the probability:** Subtract the smaller probability from the larger probability to find the probability of the random variable falling within the desired range.
P(-2.696 < Z < 2.696) = P(Z < 2.696) - P(Z < -2.696)
P(-2.696 < Z < 2.696) = 0.9965 - 0.0035 = 0.993
**Therefore, the probability of a normal random variable falling within 1.5 interquartile ranges from the population quartiles is approximately 0.993.**
**Key Points**
* The IQR is a measure of spread that is less sensitive to outliers than the standard deviation.
* In a normal distribution, about 68% of the data falls within one standard deviation of the mean, and about 95% falls within two standard deviations. Our result shows that almost all (99.3%) of the data falls within 1.5 IQRs of the quartiles. This is because the range we are considering is wider than two standard deviations.
Let me know if you have any other questions!
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