Question 1191903: for a normal distribution if the value of mean is 70 and standard deviation is 15.75 find the points
(a) in which it contains 98% area between them
(b) in which it contains 95% area between them
(c) in which it contains 85% area between them
(d) in which 60% area between them
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to find the points for the given normal distribution:
**Understanding Confidence Intervals**
A confidence interval represents a range of values within which a certain percentage of data is expected to fall. In a normal distribution, these intervals are symmetric around the mean. We use z-scores to determine the boundaries of these intervals.
**Calculations**
The formula for calculating the confidence interval is:
* Lower Bound = Mean - (Z-score * Standard Deviation)
* Upper Bound = Mean + (Z-score * Standard Deviation)
Where the Z-score corresponds to the desired confidence level. Here's a breakdown for each percentage:
**(a) 98% Area**
* Z-score for 98% (two-tailed): Approximately 2.33 (You can find this using a Z-table or a statistical calculator).
* Lower Bound: 70 - (2.33 * 15.75) ≈ 33.36
* Upper Bound: 70 + (2.33 * 15.75) ≈ 106.64
**(b) 95% Area**
* Z-score for 95% (two-tailed): Approximately 1.96
* Lower Bound: 70 - (1.96 * 15.75) ≈ 39.13
* Upper Bound: 70 + (1.96 * 15.75) ≈ 100.87
**(c) 85% Area**
* Z-score for 85% (two-tailed): Approximately 1.44
* Lower Bound: 70 - (1.44 * 15.75) ≈ 47.33
* Upper Bound: 70 + (1.44 * 15.75) ≈ 92.67
**(d) 60% Area**
* Z-score for 60% (two-tailed): Approximately 0.84
* Lower Bound: 70 - (0.84 * 15.75) ≈ 56.74
* Upper Bound: 70 + (0.84 * 15.75) ≈ 83.26
**Summary of Results**
* **98%:** The interval containing 98% of the area is approximately (33.36, 106.64).
* **95%:** The interval containing 95% of the area is approximately (39.13, 100.87).
* **85%:** The interval containing 85% of the area is approximately (47.33, 92.67).
* **60%:** The interval containing 60% of the area is approximately (56.74, 83.26).
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