Use the Cumulative Normal Distribution Table to find the z-score for which the area to its left is 0.69.The z-score for the given area is .
Question
Use the Cumulative Normal Distribution Table to find the z-score for which the area to its left is 0.69.
The z-score for the given area is .
Solution
To find the z-score for which the area to its left is 0.69 using the Cumulative Normal Distribution Table, follow these steps:
-
Look for the closest probability to 0.69 in the body of the table. This represents the cumulative probability from the mean up to the z-score.
-
Once you find the closest probability, look at the corresponding row and column headers to find the z-score.
-
The row header gives the first two digits and the decimal, and the column header gives the second decimal place.
For example, if the closest probability to 0.69 is found at the intersection of row 0.5 and column 0.04, the corresponding z-score would be 0.54.
Please note that the exact z-score for 0.69 might not be found depending on the level of precision of your table. In this case, you would need to find the two z-scores that the actual value would fall between.
Remember that the z-score represents the number of standard deviations a value is from the mean in a standard normal distribution.
Similar Questions
ind the area of the standard normal distribution that lies to the left of Z = −1.50.
Determine the area under the standard normal curve that lies between -1.98 to +0.26 (z value).Group of answer choices0.578720.397430.421280.023852
Using the standard normal curve, the area between z = 0 and z = 3.50 is about 0.50.Group of answer choicesTrueFalse
Find the Z-value with an area of 0.7134 under the standard normal curve to its right.
Suppose Z follows the standard normal distribution. Use your z table, to determine the value of c so that the following is true.2 points0.10.30.50.7
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.