using a mean of 80 and a standard deviation of 12. a) what is the score that divides the distribution such that 99% of the area is below it. b) what are the scores that bound the middle 95% of the distribution
use the same link as in your above post, except now you enter .95 and click "between"
To find the score that divides the distribution such that 99% of the area is below it, we need to use the concept of z-scores. A z-score represents the number of standard deviations a particular value is from the mean.
a) To find the score that divides the distribution such that 99% of the area is below it, we need to find the corresponding z-score using the standard normal distribution table (also known as the z-table).
First, we need to find the z-score corresponding to the cumulative probability of 0.99. By looking up this value in the z-table, we can find that the z-score is approximately 2.33.
Next, we can use the formula for z-scores to find the raw score:
z = (X - μ) / σ
Where:
- X is the raw score
- μ is the mean
- σ is the standard deviation
Rearranging the formula, we can solve for X:
X = z * σ + μ
Plugging in the values we have:
X = 2.33 * 12 + 80
Calculating this, we find:
X ≈ 105.96
Therefore, the score that divides the distribution such that 99% of the area is below it is approximately 106.
b) To find the scores that bound the middle 95% of the distribution, we need to find the corresponding z-scores.
The middle 95% of the distribution corresponds to the area between the 2.5th percentile and the 97.5th percentile. By looking up these values in the z-table, we can find that the z-scores are approximately -1.96 and 1.96, respectively.
Using the same formula as above, we can find the raw scores corresponding to these z-scores:
Lower Bound:
X = -1.96 * 12 + 80
Calculating this, we find:
X ≈ 56.48
Upper Bound:
X = 1.96 * 12 + 80
Calculating this, we find:
X ≈ 103.52
Therefore, the scores that bound the middle 95% of the distribution are approximately 56.48 and 103.52.