assume that a set of test scores is normally distributed with a mean of 100 and a standard devaiton of 20. Use the 68-95-99.7 rule to find the following quantities:
a. percentage of scores greater than 120
would the percentage of scores greater than 120 be 4.865 or 4.87
Z = (120-100)/20 = 20/20 = +1
Of the 50% higher than the mean, Z = +1 gives you
50 - 1/2(68) = 50 - 34 =?
16
To find the percentage of scores greater than 120, we can use the 68-95-99.7 rule which tells us that within one standard deviation of the mean (68%), there will be 34% of scores on each side of the mean.
First, we need to find the z-score for the value of 120 using the formula:
z = (x - μ) / σ
Where:
x = the value we want to find the percentage for (120)
μ = the mean (100)
σ = the standard deviation (20)
z = (120 - 100) / 20
z = 20 / 20
z = 1
Now that we have the z-score, we need to find the percentage of scores greater than 1 standard deviation above the mean.
According to the 68-95-99.7 rule, 34% of scores fall within one standard deviation below the mean (-1σ to +1σ). Therefore, to find the percentage of scores greater than 1 standard deviation above the mean, we subtract this percentage from 100%:
Percentage = 100% - 34%
Percentage = 66%
Therefore, the percentage of scores greater than 120 is approximately 66%.