Question

A normal distribution has a mean of 103 and a standard deviation of 10. Find the probability that a value selected at random is in the following interval.
at most 123

Answer 1

To find the probability that a randomly selected value from a normal distribution is at most 123, we can use the cumulative distribution function (CDF) of the normal distribution.

Given:
- Mean (\( \mu \)) = 103
- Standard deviation (\( \sigma \)) = 10
- Value (\( x \)) = 123

We first need to standardize the value of 123 using the z-score formula:

\[
z = \frac{x - \mu}{\sigma}
\]

Substituting the known values:

\[
z = \frac{123 - 103}{10} = \frac{20}{10} = 2
\]

Now we need to find the probability that \( Z \leq 2 \) using the standard normal distribution table, or a calculator/statistical software.

The cumulative probability for \( Z \leq 2 \) is approximately:

\[
P(Z \leq 2) \approx 0.9772
\]

Thus, the probability that a randomly selected value from this normal distribution is at most 123 is approximately 0.9772, or 97.72%.