Questions: Determine the area under the standard normal curve that lies to the left of (a) Z=1.53, (b) Z=1.43, (c) Z=0.33, and (d) Z=-1.48. (a) The area to the left of Z=1.53 is . (Round to four decimal places as needed.) (b) The area to the left of Z=1.43 is (Round to four decimal places as needed.) (c) The area to the left of Z=0.33 is (Round to four decimal places as needed.) (d) The area to the left of Z=-1.48 is (Round to four decimal places as needed.)

Determine the area under the standard normal curve that lies to the left of (a) Z=1.53, (b) Z=1.43, (c) Z=0.33, and (d) Z=-1.48.

(a) The area to the left of Z=1.53 is .
(Round to four decimal places as needed.)
(b) The area to the left of Z=1.43 is
(Round to four decimal places as needed.)
(c) The area to the left of Z=0.33 is
(Round to four decimal places as needed.)
(d) The area to the left of Z=-1.48 is
(Round to four decimal places as needed.)

Transcript text: Determine the area under the standard normal curve that lies to the left of (a) $Z=1.53$, (b) $Z=1.43$, (c) $Z=0.33$, and (d) $Z=-1.48$. \# Click the icon to view a table of areas under the normal curve. (a) The area to the left of $Z=1.53$ is $\square$ . (Round to four decimal places as needed.) (b) The area to the left of $\mathrm{Z}=1.43$ is $\square$ (Round to four decimal places as needed.) (c) The area to the left of $\mathrm{Z}=0.33$ is $\square$ (Round to four decimal places as needed.) (d) The area to the left of $Z=-1.48$ is $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Area Calculation for $ Z = 1.53 $

To find the area under the standard normal curve to the left of $ Z = 1.53 $, we use the cumulative distribution function $ \Phi(Z) $: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.53) - \Phi(-\infty) \] Since $ \Phi(-\infty) = 0 $, we have: \[ P = \Phi(1.53) = 0.937 \] Thus, the area to the left of $ Z = 1.53 $ is $ 0.937 $.

Step 2: Area Calculation for $ Z = 1.43 $

Next, we calculate the area to the left of $ Z = 1.43 $: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.43) - \Phi(-\infty) \] Again, since $ \Phi(-\infty) = 0 $: \[ P = \Phi(1.43) = 0.9236 \] Thus, the area to the left of $ Z = 1.43 $ is $ 0.9236 $.

Step 3: Area Calculation for $ Z = 0.33 $

Finally, we find the area to the left of $ Z = 0.33 $: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.33) - \Phi(-\infty) \] Again, since $ \Phi(-\infty) = 0 $: \[ P = \Phi(0.33) = 0.6293 \] Thus, the area to the left of $ Z = 0.33 $ is $ 0.6293 $.