Questions: Example: The total time spent by visitors to the Renzie Park Art Exhibit is normally distributed with a mean μ=58 minutes and a standard deviation of σ=9 minutes. What is the probability that a randomly selected visitor spends between 58 and 67 minutes at the exhibit?

Solution Steps

To find the probability that a randomly selected visitor spends between 58 and 67 minutes at the exhibit, we first calculate the Z-scores for the bounds of the range.

The Z-score for the lower bound (58 minutes) is calculated as: \[ Z_{start} = \frac{58 - \mu}{\sigma} = \frac{58 - 58}{9} = 0.0 \]

The Z-score for the upper bound (67 minutes) is calculated as: \[ Z_{end} = \frac{67 - \mu}{\sigma} = \frac{67 - 58}{9} = 1.0 \]

Next, we use the Z-scores to find the probabilities from the standard normal distribution.

The probability that a visitor spends between 58 and 67 minutes is given by: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.0) - \Phi(0.0) \]

Using standard normal distribution tables or a calculator, we find: \[ \Phi(1.0) \approx 0.8413 \quad \text{and} \quad \Phi(0.0) = 0.5 \]

Thus, the probability is: \[ P = 0.8413 - 0.5 = 0.3413 \]

Final Answer