If a set of observations is normally distributed with mean μ and variance σ², what percent of these differ from the mean by more than 1.3σ?

Steps to Solve

1. Understanding the normal distribution
   The question involves a normally distributed dataset, characterized by its mean ($\mu$) and variance ($\sigma^2$). The standard deviation ($\sigma$) is the square root of variance.

2. Determining the Z-scores
   To find the percentage of observations that differ from the mean by more than 1.3 standard deviations, we can calculate the Z-scores. The Z-score is calculated using the formula: $$ Z = \frac{X - \mu}{\sigma} $$
   For this problem, we are interested in $Z = \pm 1.3$, which means $X$ can be greater than $\mu + 1.3\sigma$ or less than $\mu - 1.3\sigma$.

3. Using the standard normal distribution table
   Look up the Z-scores in a standard normal distribution (Z) table. The table provides the area to the left of a given Z-score. For $Z = 1.3$, the corresponding area is approximately 0.9032.

4. Calculating the tails
   Since we want the area in both tails (greater than 1.3 and less than -1.3), we need to find: $$ P(Z > 1.3) + P(Z < -1.3) $$
   This can be found by calculating: $$ P(Z < -1.3) = 1 - P(Z < 1.3) $$ $$ P(Z > 1.3) = 1 - 0.9032 $$

5. Final percentage calculation
   Now calculate both tails:

• For $Z > 1.3$: $$ P(Z > 1.3) \approx 1 - 0.9032 = 0.0968 $$
• For $Z < -1.3$: $$ P(Z < -1.3) \approx 0.0968 $$
  Adding those together, the total probability is: $$ P(Z < -1.3) + P(Z > 1.3) = 0.0968 + 0.0968 = 0.1936 $$
  To convert to a percentage, multiply by 100: $$ 0.1936 \times 100 \approx 19.36% $$