On standard IQ tests, the mean is 100 and the standard deviation is 13. The results are very close to fitting a normal curve. Suppose an IQ test is given to a very large group of people. Find the percent of people whose IQ score is between 61 and 139.
First of all, on standard IQ tests (e.g., WISC and WAIS), SD = 15, but I will go with your data.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion between the two Z scores. Change to percentage.
136
To find the percent of people whose IQ score is between 61 and 139, we need to calculate the z-scores for these two values and then find the area under the normal curve between these z-scores.
First, let's find the z-score for an IQ score of 61:
z = (X - μ) / σ
where X is the IQ score, μ is the mean, and σ is the standard deviation.
z = (61 - 100) / 13
z = -39 / 13
z = -3
Next, let's find the z-score for an IQ score of 139:
z = (X - μ) / σ
z = (139 - 100) / 13
z = 39 / 13
z = 3
Now we need to find the area under the normal curve between these two z-scores, which represents the percent of people whose IQ score is between 61 and 139. We can use a standard normal distribution table or a calculator to find this area.
Using a standard normal distribution table, the area to the left of z = -3 is 0.0013, and the area to the left of z = 3 is 0.9987. To find the area between these two z-scores, we subtract the smaller area from the larger area:
Area between z = -3 and z = 3 = 0.9987 - 0.0013 = 0.9974
So, approximately 99.74% of people will have an IQ score between 61 and 139.
To find the percentage of people whose IQ score is between 61 and 139, we need to calculate the area under the normal curve between these two IQ scores.
First, let's convert the given scores into z-scores. A z-score measures the number of standard deviations a given value is from the mean. We can calculate the z-scores using the formula:
z = (x - μ) / σ
where
x is the IQ score,
μ is the mean (100),
and σ is the standard deviation (13).
For an IQ of 61:
z1 = (61 - 100) / 13
= -39 / 13
≈ -3
For an IQ of 139:
z2 = (139 - 100) / 13
= 39 / 13
≈ 3
Next, we need to find the proportion of the population that falls between these two z-scores. We can do this by looking up the z-scores in a standard normal distribution table or by using a calculator.
Using a standard normal distribution table, you can find the proportion directly. The table provides the area under the curve to the left of a given z-score. Since we want the area between z1 and z2, we need to find the area to the left of z2 and subtract the area to the left of z1.
Looking up z = 3 in the standard normal distribution table, we find an area of approximately 0.9987 to the left. Similarly, looking up z = -3, we find an area of approximately 0.0013 to the left.
To find the area between z1 and z2:
Area = Area to the left of z2 - Area to the left of z1
= 0.9987 - 0.0013
= 0.9974
Finally, we convert the proportion to a percentage by multiplying by 100:
Percentage = Area * 100
= 0.9974 * 100
≈ 99.74%
Therefore, approximately 99.74% of people would have an IQ score between 61 and 139 in a very large group of people.