scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a randomly chosen person's IQ score will be between 62 and 106, to the nearest thousandth?
To find the probability that a randomly chosen person's IQ score will be between 62 and 106, we need to calculate the z-scores for each of these values and then use the standard normal distribution table (or a calculator) to find the area under the curve between these two z-scores.
First, let's find the z-score for 62:
z1 = (X - μ) / σ
z1 = (62 - 100) / 15
z1 = -2.5333
Next, let's find the z-score for 106:
z2 = (106 - 100) / 15
z2 = 0.4
Now, we can use a standard normal distribution table or a calculator to find the area under the curve between these two z-scores.
The area to the left of z1 is 0.0056
The area to the left of z2 is 0.6554
Therefore, the probability that a randomly chosen person's IQ score will be between 62 and 106 is:
P(62 < X < 106) = P(z1 < Z < z2) = P(z < z2) - P(z < z1)
P(62 < X < 106) = 0.6554 - 0.0056
P(62 < X < 106) = 0.6498
So, the probability that a randomly chosen person's IQ score will be between 62 and 106 is approximately 0.650 (rounded to the nearest thousandth).