Math Problem Statement

Normal Distribution: Consider a normally distributed random variable with mean 0 and standard deviation 1. How does the probability that fall within the interval [−1,1] relate to probability that fall within the interval [0,1] ?

Solution

For a normally distributed random variable XX with mean μ=0\mu = 0 and standard deviation σ=1\sigma = 1, the probability that XX falls within certain intervals can be related to the properties of the standard normal distribution (denoted as ZN(0,1)Z \sim N(0, 1)).

1. Probability that XX falls within the interval [1,1][-1, 1]:

We need to compute the probability P(1X1)P(-1 \leq X \leq 1), which is the area under the standard normal curve from -1 to 1:

P(1X1)=P(Z1)P(Z1)P(-1 \leq X \leq 1) = P(Z \leq 1) - P(Z \leq -1)

Using the cumulative distribution function (CDF) for a standard normal distribution, we have:

P(Z1)0.8413andP(Z1)0.1587P(Z \leq 1) \approx 0.8413 \quad \text{and} \quad P(Z \leq -1) \approx 0.1587

So:

P(1Z1)=0.84130.1587=0.6826P(-1 \leq Z \leq 1) = 0.8413 - 0.1587 = 0.6826

This means approximately 68.26% of the values of a standard normal variable fall within one standard deviation of the mean (between -1 and 1).

2. Probability that XX falls within the interval [0,1][0, 1]:

Now, we compute P(0X1)P(0 \leq X \leq 1), which is the area under the standard normal curve from 0 to 1:

P(0Z1)=P(Z1)P(Z0)P(0 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq 0)

Since P(Z1)0.8413P(Z \leq 1) \approx 0.8413 and P(Z0)=0.5P(Z \leq 0) = 0.5, we have:

P(0Z1)=0.84130.5=0.3413P(0 \leq Z \leq 1) = 0.8413 - 0.5 = 0.3413

Thus, approximately 34.13% of the values of a standard normal variable fall between 0 and 1.

3. Relationship Between the Two Probabilities:

The probability of falling within [1,1][-1, 1] is 68.26%, while the probability of falling within [0,1][0, 1] is 34.13%. Since the standard normal distribution is symmetric around zero, the probability that ZZ falls between -1 and 0 is also 34.13%, meaning that:

P(1Z1)=2×P(0Z1)P(-1 \leq Z \leq 1) = 2 \times P(0 \leq Z \leq 1)

In other words, the probability of ZZ falling within [1,1][-1, 1] is exactly twice the probability of ZZ falling within [0,1][0, 1].


Would you like any further details or explanations?

Here are five related questions to explore:

  1. What is the probability of a standard normal variable falling between -2 and 2?
  2. How does the probability change if the interval is expanded to [2,2][-2, 2]?
  3. Can you derive the probability for intervals such as [1,2][1, 2] using the CDF?
  4. What are the properties of the normal distribution that lead to symmetry?
  5. How does the concept of z-scores relate to probabilities in normal distribution?

Tip: Always remember that about 68.26%, 95.44%, and 99.73% of the data falls within 1, 2, and 3 standard deviations of the mean, respectively, in a standard normal distribution.

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Math Problem Analysis

Mathematical Concepts

Probability
Normal Distribution
Standard Normal Distribution
Symmetry of Normal Distribution

Formulas

P(-1 ≤ Z ≤ 1) = P(Z ≤ 1) - P(Z ≤ -1)
P(0 ≤ Z ≤ 1) = P(Z ≤ 1) - P(Z ≤ 0)
P(-1 ≤ Z ≤ 1) = 2 * P(0 ≤ Z ≤ 1)

Theorems

68-95-99.7 Rule (Empirical Rule)
Properties of Standard Normal Distribution

Suitable Grade Level

Grades 10-12