
In the movie Forrest Gump, the public school required an IQ of at least 80 for admittance.

(a)	
If IQ test scores are normally distributed with mean 100 and standard deviation 16, what percentage of people would qualify for admittance to the school? (Round your answer to 2 decimal places.)


  Percentage of people	
89.44
 %  

(b)	
If the public school wishes 95 percent of all children to qualify for admittance, what minimum IQ test score should be required for admittance? (Round the z value to 3 decimal places. Round your answer to 2 decimal places.

To solve both parts of this problem, we will use the properties of a normally distributed data set. Specifically, we will apply the standard normal distribution and z-scores.

### Part (a):
We are given:
- Mean IQ score (\(\mu\)) = 100
- Standard deviation (\(\sigma\)) = 16
- Minimum IQ for admittance = 80

We need to find what percentage of people would have an IQ score of 80 or higher.

#### Step 1: Calculate the z-score
The z-score formula is:

\[
z = \frac{x - \mu}{\sigma}
\]

Where:
- \(x\) is the value we're evaluating (IQ of 80 in this case),
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.

\[
z = \frac{80 - 100}{16} = \frac{-20}{16} = -1.25
\]

#### Step 2: Use the z-score table or a calculator
The z-score of -1.25 corresponds to a cumulative probability of approximately 0.1056. This means that 10.56% of people have an IQ lower than 80. Therefore, the percentage of people with an IQ of 80 or higher is:

\[
1 - 0.1056 = 0.8944 \quad \text{or} \quad 89.44\%
\]

So, **89.44%** of people would qualify for admittance.

### Part (b):
Now, we are asked to determine the minimum IQ score required for admittance if the school wants 95% of all children to qualify.

#### Step 1: Find the z-score corresponding to 95%
Since we want 95% of children to qualify, we are looking for the IQ score where 95% of the population has a lower IQ, meaning we need the z-score corresponding to the 95th percentile.

From the z-score table, the z-score corresponding to 95% is approximately **1.645**.

#### Step 2: Use the z-score formula to find the minimum IQ
Now, using the z-score formula in reverse:

\[
x = \mu + z \cdot \sigma
\]

Substituting the values:

\[
x = 100 + (1.645 \cdot 16) = 100 + 26.32 = 126.32
\]

So, the minimum IQ score should be approximately **126.32**.

### Final answers:
- **(a)**: 89.44% of people qualify for admittance.
- **(b)**: The minimum IQ score required for 95% of children to qualify is **126.32**.

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the z-score for an IQ of 120?
2. How does changing the standard deviation affect the percentage of people who qualify for admittance?
3. What is the probability of having an IQ score higher than 130?
4. How do you interpret a z-score of -2.5 in the context of IQ scores?
5. What percentile does an IQ of 110 correspond to?

**Tip:** When interpreting z-scores, a positive z-score means the value is above the mean, and a negative z-score means the value is below the mean.

In the movie Forrest Gump, the public school required an IQ of at least 80 for admittance.

(a) If IQ test scores are normally distributed with mean 100 and standard deviation 16, what percentage of people would qualify for admittance to the school? (Round your answer to 2 decimal places.)

(b) If the public school wishes 95 percent of all children to qualify for admittance, what minimum IQ test score should be required for admittance? (Round the z value to 3 decimal places. Round your answer to 2 decimal places.)

Solve a problem involving the normal distribution of IQ scores. Learn how to calculate the percentage of people qualifying for admittance to a school based on IQ, and determine the minimum IQ required for 95% of students to qualify. Key concepts include z-scores and percentiles, using a mean of 100 and a standard deviation of 16.

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