Question 5.5: Courier Service Delivery Time Study A courier service compan......

(a) Find P(45 < x < 51)

Step 1: Always sketch a normal probability distribution and indicate the area (probability) to be found, as shown in Figure 5.9.

Step 2: Transform the x-limits into corresponding z-limits using Formula 5.6.

In this example, x = 45 corresponds to z=\frac{45-45}{8} =0

and x = 51 corresponds to z=\frac{51-45}{8} =0.75.

Thus P(45 < x < 51) is equivalent to finding P(0 < z < 0.75).

Step 3: Calculate the required probability of P(0 < z < 0.75) using the z-table.

The area between (0 < z < 0.75) is equal to 0.2734. Then P(45 < x < 51) = 0.2734.

Then P(45 < x < 51) = 0.2734.

Thus there is a 27.34% chance that a randomly selected parcel will take between 45 minutes and 51 minutes to deliver to the client.

(b) Find P(x < 48)

Step 1: Sketch a normal probability distribution and indicate the area (probability) to be found, as shown in Figure 5.10.

Step 2: Transform the x-limits into corresponding z-limits using Formula 5.6.

In this example, x = 48 corresponds to z=\frac{48-45}{8} =0.375.

Thus P(x < 48) is equivalent to finding P(z < 0.375).

Step 3: Calculate the required probability P(z < 0.375) using the z-table.

P(z < 0.375) is found by splitting the area into P(z < 0) and P(0 < z < 0.375),

since the z-table only calculates the area above z = 0.

The area between (0 < z < 0.375) is equal to 0.1480 (use z = 0.38) while the area below z = 0 is equal to 0.5 (the lower half of the normal curve).

Then P(x < 48) = P(z < 0.375) = 0.5 + 0.1480 = 0.6480

Thus there is a 64.8% chance that a randomly selected parcel will be delivered to the client within 48 minutes.