Find the probability that a randomly selected adult has an IQ between 89 and 111.

Question

Assume that adults have IQ scores that are normally distributed with a mean of μ=100

and a standard deviation σ=20.

Find the probability that a randomly selected adult has an IQ between

89 and 111

Answer 1

IQ's between 89 and 111 are all within 11/20 standard deviations from the mean, or 0.55

68/% fall within 1 standard deviation. Those within 0.55 standard deviations from the mean should be

substantially less than 68% the mean minus the IQ all divided by the standard deviation = z score.

Look up a z-score of 0.55 and find the corresponding %. .2088 is across from row 0.5 and below the column .05

0.2088 = 20.88% of the IQ's are between 100 and 89, and also between 100 and 111

2 x 0.2088 = 0.4176 = 41.76 % of the IQ's are between 89 and 111

IF you randomly selected someone the probability that they had an IQ between 89 and 111 would be 41.76%

Answer 2

This question is asking how much of probability curve is located between IQ scores of 89 and 111.

Step 1: Calculate z-scores (for normal distribution).

z = (x - μ) / σ

z₁ = (111 - 100) / 20 = 11 / 20 = 0.55

z₂ = (89 - 100) / 10 = -11 / 20 = -0.55

Step 2: Identify the appropriate portion of the curve.

The portion between z₁ and z₂.

Step 3: Evaluate the probability at each z value. (You can use a z-score table or an online calculator, such as https://www.calculator.net/z-score-calculator.html)

P(z < 0.55) = 70.884%

P(z < -0.55) = 29.116%

Ans: P(z < 0.55) - P(z <= -0.55) = 70.884% - 29.116% = 41.769%

41.769%