Question

The number of vehicles leaving a highway at a certain exit during a particular time period has a distribution that is approximately normal with mean value 500 and standard deviation \(75 .\) What is the probability that the number of cars exiting during this period is a. At least \(650 ?\) b. Strictly between 400 and \(550 ?\) (Strictly means that the values 400 and 550 are not included.) c. Between 400 and 550 (inclusive)?

Short answer

a. The probability that the number of cars exiting during this period is at least 650 is calculated using the z-value and z-table. Similarly, b. the probability of cars exiting being strictly between 400 and 550, and c. between 400 and 550 inclusive, are both calculated using corresponding z-values and the z-table.

Step-by-step solution

Calculation for at least 650 cars

Firstly, find the z-value for 650 using the formula \(z = \frac{(650 - 500)}{75}\). The resulting z-value is then used to find the probability associated with it in a z-table, to find the probability \(P(X ≥ 650)\), you look up the probability for \(P(Z ≤ z)\) in the z-table and then subtract this value from 1 because the value is 'at least 650'.

Calculation for strictly between 400 and 550

Find the respective z-values for 400 and 550 using the formula \(z = \frac{(X - µ)}{σ}\). Find the probabilities associated with these z-values in a z-table. The probability \(P(400 < X < 550)\) is calculated as \(P(Z ≤ z_{550}) - P(Z ≤ z_{400})\).

Calculation for between 400 and 550 (inclusive)

The calculation is identical to step 2, since in a continuous distribution like the normal distribution, the probability at a single specific value is technically 0. Hence, the result will not change even when 400 and 550 are included. The probability for this range is \(P(400 ≤ X ≤ 550) = P(400 < X < 550) = P(Z ≤ z_{550}) - P(Z ≤ z_{400})\).