Question

"IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A psychologist wants to know the probability that a random selected individual will have an IQ score of 95 or greater. What steps would you use to solve the problem? Place four steps below in the correct order.

1. Report the probability as either a decimal or a percentage.
2. Read the probability in Column C (the tail) in the unit normal table associated with the relevant Z score.
3. Substitute the raw score, mean and standard deviation values into the z formula and calculate the z score.
4. look up the Unit Normal Table and find the relevant z score in Column A.
a. Describe the general purpose of a z score and explain how a z score accomplishes this goal.
b. Describe what happens to the mean, the standard deviation, and the shape of a distribution when all of the scores are transformed into z scores."

Answer 1

To find the probability of a randomly selected individual having an IQ score of 95 or greater, calculate the z-score using the formula (X - μ) / σ and look it up in the unit normal table to find the corresponding probability. The probability can be reported as either a decimal or percentage.

Step-by-step explanation:

To solve this problem, we need to find the probability that a randomly selected individual will have an IQ score of 95 or greater given that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.

1. Substitute the raw score, mean, and standard deviation values into the z-score formula and calculate the z-score using the formula: z = (X - μ) / σ, where X is the raw score, μ is the mean, and σ is the standard deviation.
2. Look up the z-score in the unit normal table and find the corresponding probability in Column C (the tail) of the table.
3. Report the probability as either a decimal or a percentage.

In this case, we want to find the probability of an IQ score of 95 or greater, which translates to finding the probability to the right of 95. So we calculate the z-score as follows: z = (95 - 100) / 15 = -0.3333. Looking up the z-score in the unit normal table, we find the probability to the right of -0.3333 is 0.6297. Therefore, the probability that a randomly selected individual will have an IQ score of 95 or greater is 0.6297, which can also be expressed as 62.97%.