2 Multiple choice 1 point 1Qscores are normally distributed with mean 100 and standard deviation 15 . Find the probability of an IQ score falling between 90 and 110 . 0.3187 0.5050 0.6813 0.4950

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We are given that the IQ scores are normally distributed with a mean $$ \mu = 100 $$ and a standard deviation $$ \sigma = 15 $$. We want to find

$$
P(90 \leq X \leq 110).
$$

1. **Standardize the scores:**  
   
   For $$X = 90$$:

$$
   z_1 = \frac{90 - 100}{15} = \frac{-10}{15} = -\frac{2}{3}
   $$
   
   For $$X = 110$$:

$$
   z_2 = \frac{110 - 100}{15} = \frac{10}{15} = \frac{2}{3}
   $$

2. **Express the probability in terms of $$z$$:**

$$
   P(90 \leq X \leq 110) = P\left(-\frac{2}{3} \leq Z \leq \frac{2}{3}\right)
   $$
   
   where $$Z$$ is a standard normal variable.

3. **Use symmetry of the normal distribution:**

The probability can be written as:

$$
   P\left(-\frac{2}{3} \leq Z \leq \frac{2}{3}\right) = \Phi\left(\frac{2}{3}\right) - \Phi\left(-\frac{2}{3}\right)
   $$
   
   Since

$$
   \Phi\left(-\frac{2}{3}\right) = 1 - \Phi\left(\frac{2}{3}\right),
   $$

we have:

$$
   P\left(-\frac{2}{3} \leq Z \leq \frac{2}{3}\right) = \Phi\left(\frac{2}{3}\right) - \left[1 - \Phi\left(\frac{2}{3}\right)\right] = 2\Phi\left(\frac{2}{3}\right) - 1.
   $$

4. **Approximate using the standard normal table:**

For $$ z = \frac{2}{3} \approx 0.667 $$, the standard normal cumulative probability is approximately:

$$
   \Phi\left(0.667\right) \approx 0.7486.
   $$

Therefore:

$$
   P(90 \leq X \leq 110) \approx 2(0.7486) - 1 = 1.4972 - 1 = 0.4972.
   $$

5. **Select the closest answer:**

Among the given choices, $$0.4950$$ is the closest to $$0.4972$$.

Thus, the probability that an IQ score falls between 90 and 110 is approximately $$0.4950$$.
The probability of an IQ score between 90 and 110 is approximately 0.4950.

2 Multiple choice 1 point 1Qscores are normally distributed with mean 100 and standard deviation 15 . Find the probability of an IQ score falling between 90 and 110 . 0.3187 0.5050 0.6813 0.4950

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