Question

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

Answer 1

The area of the shaded region between IQ scores of 106 and 128, in a normal distribution with a mean of 100 and a standard deviation of 15, is approximately 31.39%.

The area of the shaded region between IQ scores of 106 and 128, in a normal distribution with a mean of 100 and a standard deviation of 15, is approximately 31.39%.

To find the area of the shaded region between IQ scores of 106 and 128 in a normal distribution with a mean of 100 and a standard deviation of 15, you can use the z-score formula:

`\[ Z = ((X - \mu))/(\sigma) \]`

where:

- X is the IQ score,

-
`\( \mu \)` is the mean,

-
`\( \sigma \)` is the standard deviation.

For the lower limit (106):

`\[ Z_{\text{lower}} = ((106 - 100))/(15) = (6)/(15) = 0.4 \]`

For the upper limit (128):

`\[ Z_{\text{upper}} = ((128 - 100))/(15) = (28)/(15) = 1.87 \]`

Now, you can use a standard normal distribution table or a calculator to find the area between these two z-scores.

The area between
`\( Z_{\text{lower}} \)` and
`\( Z_{\text{upper}} \)` represents the probability that a randomly selected IQ score falls between 106 and 128.

For Z = 0.4, the probability is approximately 0.6554.

For Z = 1.87, the probability is approximately 0.9693.

Now, to find the area between these two z-scores, subtract the lower probability from the upper probability:

`\[ \text{Area} = 0.9693 - 0.6554 \]\[ \text{Area} \approx 0.3139 \]`

So, the area of the shaded region between IQ scores of 106 and 128 is approximately 0.3139, or 31.39%.