Find the probability that a randomly selected person has an IQ score between 88 and
112.
To find the probability that a randomly selected person has an IQ score between 88 and 112, you will need to know the mean and standard deviation of the IQ scores distribution. The mean and standard deviation of IQ scores are typically 100 and 15, respectively.
Next, you can use the Z-score formula to calculate the Z-scores for the given IQ scores of 88 and 112. The Z-score formula is:
Z = (X - μ) / σ
where X is the value you are interested in (88 or 112 in this case), μ is the mean, and σ is the standard deviation.
For IQ score 88:
Z1 = (88 - 100) / 15 = -0.8
For IQ score 112:
Z2 = (112 - 100) / 15 = 0.8
Now, you will need to find the corresponding probabilities from the standard normal distribution table or use a calculator to find the area under the standard normal curve between these two Z-scores. The area gives you the probability.
Alternatively, you can use technology, like a graphing calculator or statistical software, to find the probability directly. By inputting the mean, standard deviation, and the range of 88 to 112, you can obtain the probability directly from the calculator or software.
Using either method, the calculated probability will be the same.
To find the probability that a randomly selected person has an IQ score between 88 and 112, you need to know the distribution of IQ scores in the population and the mean and standard deviation of the distribution.
Assuming that IQ scores follow a normal distribution, you can use the Z-score formula to calculate the probability. The Z-score formula transforms raw scores into standard scores by subtracting the mean and dividing by the standard deviation.
1. Determine the mean and standard deviation of the IQ scores in the population. Let's say the mean is μ and the standard deviation is σ.
2. Calculate the Z-scores for the lower and upper bounds of the range you're interested in: Z1 = (88 - μ) / σ and Z2 = (112 - μ) / σ.
3. Look up the probabilities associated with each Z-score in the standard normal distribution table or use a Z-score calculator. The standard normal distribution table gives you the probability of a Z-score falling below a certain value, so you need to subtract the probability associated with the lower Z-score from the probability associated with the upper Z-score.
4. Calculate the probability by subtracting the lower Z-score probability from the upper Z-score probability: P = P(Z1 ≤ Z ≤ Z2) = P(Z ≤ Z2) - P(Z ≤ Z1).
By following these steps, you can find the probability that a randomly selected person has an IQ score between 88 and 112.
The "sigma" or standard deviation for IQ sorces is 16, and the mean is 100. You are asking for the probability that the IQ is between -0.75 sigma and +0.75 sigma of the mean of a normal distribution.
Using the computational tool at
(Broken Link Removed) I get 54.7%