Question

You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (a) The area to the left of z is 0.1841 (b) The area between -z and z is 0.9398. (c) The area between -z and z is 0.2282 (d) The area to the left of z is 0.9951. (e) The area to the right of z is 0.6554.

You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (a) The area to the left of z is 0.1841 (b) The area between -z and z is 0.9398. (c) The area between -z and z is 0.2282 (d) The area to the left of z is 0.9951. (e) The area to the right of z is 0.6554.

Answer 1

a) The area to the left of \(z\) is 0.1841

\(\mathrm{P}(Z \leq \mathrm{z})=0.1841\)

\(\mathrm{P}(-\infty < Z \leq 0)-\mathrm{P}(\mathrm{z} < \mathrm{Z} \leq 0)=0.1841\)

\(0.5-\mathrm{P}(\mathrm{z} < Z \leq 0)=0.1841\)

\(\mathrm{P}(\mathrm{z} < Z \leq 0)=0.5-0.1841\)

\(\mathrm{P}(\mathrm{z} < Z \leq 0)=0.3159\)

From z table, the z value is \(-\mathbf{0 . 9 0}\)

Thus, \(z=-0.90\)

b) The area between \(-z\) and \(z\) is 0.9398

\(\mathrm{P}(-\mathrm{z} \leq \mathrm{Z} \leq \mathrm{z})=0.9398\)

\(\mathrm{P}(-\mathrm{z} \leq \mathrm{Z} \leq 0)+\mathrm{P}(0 \leq \mathrm{Z} \leq \mathrm{z})=0.9398\)

\(2 * \mathrm{P}(0 \leq Z \leq \mathrm{z})=0.9398\)

\(\mathrm{P}(0 \leq Z \leq \mathrm{z})=0.9398 / 2\)

\(\mathrm{P}(0 \leq Z \leq z)=0.4699\)

From z table, the z value is 1.88

Thus, \(z=1.88\)

c) The area between \(-z\) and \(z\) is 0.2282

\(\mathrm{P}(-\mathrm{z} \leq \mathrm{Z} \leq \mathrm{z})=0.2282\)

\(\mathrm{P}(-\mathrm{z} \leq \mathrm{Z} \leq 0)+\mathrm{P}(0 \leq \mathrm{Z} \leq \mathrm{z})=0.2282\)

\(2 * \mathrm{P}(0 \leq Z \leq z)=0.2282\)

\(\mathrm{P}(0 \leq \mathrm{Z} \leq \mathrm{z})=0.2282 / 2\)

\(\mathrm{P}(0 \leq Z \leq \mathrm{z})=0.1141\)

From z table, the z value is 0.29

Thus, \(z=0.29\)

d) The area to the left of \(z\) is 0.9951

\(\mathrm{P}(Z \leq \mathrm{z})=0.9951\)

\(\mathrm{P}(-\infty < Z \leq 0)+\mathrm{P}(0 \leq Z \leq \mathrm{z})=0.9951\)

\(0.5+\mathrm{P}(0 \leq Z \leq \mathrm{z})=0.9951\)

\(\mathrm{P}(0 \leq Z \leq \mathrm{z})=0.9951-0.5\)

\(\mathrm{P}(0 \leq Z \leq \mathrm{z})=0.4951\)

From z table, the z value is 2.58

Thus, \(z=2.58\)

e) The area to the right of \(z\) is 0.6554

\(\mathrm{P}(Z \geq \mathrm{z})=0.6554\)

\(\mathrm{P}(\mathrm{z} \leq Z \leq 0)+\mathrm{P}(0 \leq Z<\infty)=0.6554\)

\(\mathrm{P}(\mathrm{z} \leq Z \leq 0)+0.5=0.6554\)

\(\mathrm{P}(\mathrm{z} \leq Z \leq 0)=0.6554-0.5\)

\(\mathrm{P}(\mathrm{z} \leq Z \leq 0)=0.1554\)

From z table, the z value is \(-\mathbf{0 . 4 0}\)

Thus, \(z=-0.40\)