On standard IQ tests, the mean is 100 and the standard deviation is 16. 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 more than 100.
To find the percent of people whose IQ score is more than 100, we need to use the properties of the normal distribution.
1. Calculate the z-score:
The z-score measures the number of standard deviations away from the mean a particular value is. It is calculated using the formula:
z = (x - μ) / σ
Where:
x = the value we are interested in
μ = the mean of the distribution (100 in this case)
σ = the standard deviation of the distribution (16 in this case)
In this case, we want to find the percent of people whose IQ score is more than 100. Since the mean is 100, we plug in x = 100 in the formula:
z = (100 - 100) / 16 = 0/16 = 0
2. Lookup the z-score in the z-table:
The z-table provides the cumulative probability for a given z-score. We are interested in finding the percent of people whose IQ score is above the mean, which corresponds to finding the area to the right of the z-score.
Using the z-table, we look up the value for 0. The value obtained is 0.5, representing 50% of the distribution.
3. Calculate the percentage:
Since we are interested in the percentage of people whose IQ score is more than 100, we subtract the cumulative probability value obtained in step 2 from 1 (to get the remaining area to the right of the z-score).
Percentage = (1 - 0.5) * 100 = 0.5 * 100 = 50%
Therefore, the percent of people whose IQ score is more than 100 is 50%.
To find the percent of people whose IQ score is more than 100, we need to calculate the area under the normal curve to the right of the IQ score of 100.
Step 1: Standardize the value:
We will convert the IQ score 100 to a standardized score, also known as a z-score. The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = The IQ score
μ = The mean IQ score
σ = The standard deviation of IQ scores
In this case,
x = 100
μ = 100
σ = 16
Plugging in the values:
z = (100 - 100) / 16
z = 0 / 16
z = 0
So, the standardized score (z-score) for the IQ score of 100 is 0.
Step 2: Find the area under the normal curve:
Now, we will find the area to the right of the z-score 0. This represents the percentage of people whose IQ scores are more than 100.
Using a standard normal distribution table (also known as a Z-table) or statistical software, we can find that the area to the right of the z-score 0 is approximately 0.5 or 50%.
Step 3: Convert to percentage:
Since we are looking for a percentage, we need to convert the decimal value to a percentage.
The percent of people whose IQ score is more than 100 is 50%.
Therefore, approximately 50% of the people would have an IQ score greater than 100.