Question 5.4: Using the z-table to Find Normal Probabilities Refer to the ......

(a) Refer to Figure 5.3, which shows the required area.

From the z-table, identify z = 1.4 down the left column of z. Then read across the columns at the z = 1.4 row until z = 0.06. The area found at this intersection is 0.4279. So P(0 < z < 1.46) = 0.4279

Thus there is a 42.79% chance that a z-value will lie between z = 0 and z = 1.46.

(b) Refer to Figure 5.4, which shows the required area.

Because of symmetry, this area (probability) is equivalent to finding the area P(0 < z < 2.3) on the positive z side. Reading from the z-table (left column), at z = 2.3, this area equals 0.48928.

So P(−2.3 < z < 0) = 0.48928

Thus there is a 48.93% chance that a z-value will lie between z = −2.3 and z = 0.

(c) Refer to Figure 5.5, which shows the required area.

The z-table only gives areas between the midpoint (z = 0) and an upper z-limit, k (i.e. between 0 < z < k). The area above k is found by subtracting the area between (0 < z < k) from 0.5, which is the total area above the midpoint of z = 0.

From the z-table, P(0 < z < 1.82) = 0.4656.

Then P(z > 1.82) = 0.5000 – 0.4656 (Complementary probability rule.) = 0.0344

Thus there is only a 3.44% chance that a z-value will lie above z = 1.82.

(d) Refer to Figure 5.6, which shows the required area.

P(−2.1 < z < 1.32) is equal to the sum of two mutually exclusive areas, each of which can be looked up separately in the z-table.

Thus P(−2.1 < z < 1.32) = P(−2.1 < z < 0) + P(0 < z < 1.32)

From the z-table         P(−2.1 < z < 0) = 0.4821 (Using symmetry.)

and         P(0 < z < 1.32) = 0.4066

Then P(−2.1 < z < 1.32) = 0.4821 + 0.4066 = 0.8887

Thus there is an 88.87% chance that a z-value will lie between z = −2.1 and z = 1.32.

(e) Refer to Figure 5.7, which shows the required area.

Again, split the required area into two parts, P(0 < z < 1.24) and P(0 < z < 2.075).

The difference between these two areas isolates the required probability.

From the z-table P(0 < z < 2.075) = 0.4812      (Note: 2.075 is rounded to 2.08.)

P(0 < z < 1.24) = 0.3925

Then P(1.24 < z < 2.075) = 0.4812 – 0.3925 = 0.0887

Thus there is only an 8.87% chance that a z-value lies between z = 1.24 and z = 2.075.