Question # Determine the area under the standard normal curve that lies between ​ (a) Upper Z equals -2.03 and Upper Z equals 2.03​, ​(b) Upper Z equals -1.56 and Upper Z equals 0​, and ​(c) Upper Z equals -1.51 and Upper Z equals 0.68. ​ ​(Round to four decimal places as​ needed.)

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ANSWERED Determine the area under the standard normal curve that lies between ​
(a) Upper Z equals -2.03 and Upper Z equals 2.03​,
​(b) Upper Z equals -1.56 and Upper Z equals 0​, and
​(c) Upper Z equals -1.51 and Upper Z equals 0.68. ​ ​(Round to four decimal places as​ needed.) 2021-02-09

(a) Given:
$$\displaystyle-{2.03}<{z}<{2.03}$$
Determine the corresponding probability using the normal probability table in the appendix.
$$\displaystyle{P}{\left({Z}<-{2.03}\right)}$$ is given in the row starting with -2.0 and in the column starting with .03 of the standard normal probability table in the appendix.
$$\displaystyle{P}{\left({Z}<{2.03}\right)}$$ is given in the row starting with 2.0 and in the column starting with .03 of the standard normal probability table in the appendix.
The probability between two boundaries is then the difference between the probabilities to the left of the boundaries.
$$\displaystyle{P}{\left(-{2.03}<{z}<{2.03}\right)}={P}{\left({z}<{2.03}\right)}-{P}{\left({z}<-{2.03}\right)}$$
$$\displaystyle={0.9788}-{0.0212}$$
$$\displaystyle={0.9576}$$
Thus the area under the normal distribution between $$\displaystyle-{2.03}{\quad\text{and}\quad}{2.03}$$ is approximately 0.9576.
(b) Given:
$$\displaystyle-{156}<{2}{z}<{0}$$
Determine the corresponding probability using the normal probability table in the appendix.
$$\displaystyle{P}{\left({Z}<-{1.56}\right)}$$ is given in the row starting with -1.5 and in the column starting with .06 of the standard normal probability table in the appendix.
$$\displaystyle{P}{\left({Z}<{0}\right)}$$ is given in the row starting with 0.0 and in the column starting with .00 of the standard normal probability table in the appendix.
The probability between two boundaries is then the difference between the probabilities to the left of the boundaries.
$$\displaystyle{P}{\left(-{1.56}<{z}<{0}\right)}={P}{\left({z}<{0}\right)}-{P}{\left({z}<-{1.56}\right)}$$
$$\displaystyle={0.5000}-{0.0594}$$
$$\displaystyle={0.4406}$$
Thus the area under the normal distribution between -1.56 and 0 is approximately 0.4406.
(c) Given:
$$\displaystyle-{1.51}<{z}<{0.68}$$
Determine the corresponding probability using the normal probability table in the appendix.
$$\displaystyle{P}{\left({Z}<-{1.51}\right)}$$ is given in the row starting with -1.5 and in the column starting with .01 of the standard normal probability table in the appendix.
$$\displaystyle{P}{\left({Z}<{0.68}\right)}$$ is given in the row starting with 0.6 and in the column starting with .08 of the standard normal probability table in the appendix.
The probability between two boundaries is then the difference between the probabilities to the left of the boundaries.
$$\displaystyle{P}{\left(-{1.51}<{z}<{0.68}\right)}={P}{\left({z}<{0.68}\right)}-{P}{\left({z}<-{1.51}\right)}$$
$$\displaystyle={0.7517}-{0.0655}$$
$$\displaystyle={0.6862}$$
Thus the area under the normal distribution between -1.51 and 0.68 is approximately 0.6862.