If a set of observations is normally distributed with mean μ and variance σ², what percent of these differ from the mean by less than 0.52σ?

Steps to Solve

1. Identify the standard deviation range We need to determine the range of values that differ from the mean by less than $0.52\sigma$. This range can be represented as: $$ \mu - 0.52\sigma < X < \mu + 0.52\sigma $$

2. Convert to z-scores To find the percentage of observations within this range, we convert the limits to z-scores using the formula: $$ z = \frac{X - \mu}{\sigma} $$ Thus, the z-scores corresponding to $X = \mu - 0.52\sigma$ and $X = \mu + 0.52\sigma$ are: $$ z_1 = \frac{\mu - 0.52\sigma - \mu}{\sigma} = -0.52 $$ $$ z_2 = \frac{\mu + 0.52\sigma - \mu}{\sigma} = 0.52 $$

3. Look up z-scores in standard normal distribution tables Using a standard normal distribution table, we find the probabilities corresponding to $z_1 = -0.52$ and $z_2 = 0.52$.

4. Calculate the percentage difference The area under the normal curve between these two z-scores gives us the percentage of observations that fall within this range. The percentage can be calculated as: $$ P(-0.52 < Z < 0.52) = P(Z < 0.52) - P(Z < -0.52) $$

5. Use z-table values From standard normal distribution tables:

• For $z = -0.52$, the area is approximately 0.3015.
• For $z = 0.52$, the area is approximately 0.7019.

Thus: $$ P(-0.52 < Z < 0.52) = 0.7019 - 0.3015 = 0.4004 $$

6. Convert to percentage To express this as a percentage, multiply by 100: $$ 0.4004 \times 100 \approx 40.04% $$