The Intelligence Quotient (IQ) test scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a person would score 130 or more on the test?
Using the normal distribution calculation tool at http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html ,
I get 2.28% for the answer.
Enter 100 for the mean and 15 for the std dev. Then enter 130 for start and a very high number for which the probability is negligible (like 200 or more) for the end, and hit the "enter " key.
To find the probability that a person would score 130 or more on the test, we need to determine the area under the normal distribution curve to the right of 130.
Step 1: Calculate the z-score for the value 130.
The z-score formula is: z = (x - μ) / σ
Where:
x = The value we are interested in (130 in this case)
μ = The mean of the distribution (100 in this case)
σ = The standard deviation of the distribution (15 in this case)
Plugging in the values:
z = (130 - 100) / 15
z = 2
Step 2: Use a standard normal table or calculator to find the area to the right of z = 2. This represents the probability of scoring 130 or more on the test.
Looking up the value in a standard normal table or using a calculator, we find that the area to the right of z = 2 is approximately 0.0228.
Therefore, the probability of a person scoring 130 or more on the test is approximately 0.0228, or 2.28%.
To find the probability that a person would score 130 or more on the IQ test, we need to calculate the z-score for this particular score.
The z-score is a measure of how many standard deviations a given value is from the mean. It can be calculated using the formula:
z = (x - μ) / σ,
where x is the score, μ is the mean, and σ is the standard deviation.
In this case, x = 130, μ = 100, and σ = 15.
Plugging these values into the formula, we get:
z = (130 - 100) / 15,
= 2.
Next, we need to find the probability associated with this z-score.
Using a standard normal distribution table or a calculator, we can find that the probability associated with a z-score of 2 is approximately 0.9772.
Therefore, the probability that a person would score 130 or more on the IQ test is approximately 0.9772, or 97.72%.