Assume Z has a standard normal distribution. Use Appendix Table III to determine the value for z that solves each of the following:
(Round your answer to 2 decimal places.)
P(Z<z)=.9
P(Z<z)=.5
We do not have access to Table III.
To determine the value for z that solves each of the given probabilities, we need to use the Standard Normal Distribution table (also known as Z-table or Appendix Table III).
1. P(Z < z) = 0.9:
To find the value of z that corresponds to a cumulative probability of 0.9, we need to locate the corresponding row and column in the Z-table.
- First, locate the row that contains the first digit of the probability, which is 0.9. In this case, the row is 1.2.
- Next, locate the column that contains the second digit of the probability, which is 0.0. In this case, the column is 0.09.
The intersection of row 1.2 and column 0.09 in the Z-table gives us the value of z. Reading the value from the table, we find that z is approximately 1.28.
Therefore, the value of z that solves P(Z < z) = 0.9 is approximately 1.28.
2. P(Z < z) = 0.5:
To find the value of z that corresponds to a cumulative probability of 0.5, we follow the same steps as for the previous example.
- Locate the row that contains the first digit of the probability, which is 0.5. In this case, the row is 0.6.
- Locate the column that contains the second digit of the probability, which is 0.0. In this case, the column is 0.00.
The intersection of row 0.6 and column 0.00 in the Z-table gives us the value of z. Reading the value from the table, we find that z is approximately 0.00.
Therefore, the value of z that solves P(Z < z) = 0.5 is approximately 0.00.
Remember to round the values to 2 decimal places as specified in the question.
To find the value of z that solves each of the given probabilities, we can use the z-table (Appendix Table III).
1. P(Z < z) = 0.9:
Looking at the z-table, find the value closest to 0.9 in the middle column. The corresponding z-value is 1.28.
2. P(Z < z) = 0.5:
Locate the value closest to 0.5 in the middle column of the z-table. The corresponding z-value is 0.00 (or 0 if rounded to two decimal places).
So, the values of z that solve each of the given probabilities are:
1. z = 1.28
2. z = 0.00 (or 0)