Final answer:
To find the probabilities and z-values for a standard normal random variable:
(a) P(z<-1.72) is approximately 0.0429.
(b) P(z>0.75) is approximately 0.2266.
(c) P(-1.25<z<0.37) is approximately 0.5387.
(d) The z-value with an area to the right of 0.38 is approximately 0.26.
(e) The z-value with an area to the left of 0.44 is approximately 0.44.
(f) The area between -z and 0 is 0.5.
Step-by-step explanation:
(a) To find P(z<-1.72), we look up the corresponding area to the left of -1.72 in the z-table. The z-table shows that the area to the left of -1.72 is approximately 0.0429.
(b) To find P(z>0.75), we need to find the area to the right of 0.75. The z-table shows that the area to the left of 0.75 is approximately 0.7734. Therefore, the area to the right of 0.75 is 1 - 0.7734 = 0.2266.
(c) To find P(-1.25<z<0.37), we need to find the area between these two z-values. The z-table shows that the area to the left of -1.25 is approximately 0.1056, and the area to the left of 0.37 is approximately 0.6443. Therefore, the area between them is 0.6443 - 0.1056 = 0.5387.
(d) To find the z-value with an area to the right of 0.38, subtract 0.38 from 1 to get 0.62. Looking up this area in the z-table, we find that the corresponding z-value is approximately 0.26.
(e) To find the z-value with an area to the left of 0.44, simply look up this area in the z-table. The z-table shows that the area to the left of 0.44 is approximately 0.6700. Therefore, the corresponding z-value is approximately 0.44.
(f) To find the area between -z and 0, we need to find the area to the left of -z and to the left of 0, and then subtract the smaller area from the larger area. The z-table shows that the area to the left of -z is 0.5 - z/2, and the area to the left of 0 is 0.5. Setting these two equal, we get 0.5 - z/2 = 0.5, which simplifies to z = 0. Therefore, the area between -z and 0 is 0.5 - 0 = 0.5.