Question 1192634
**a. Percentage of people who would qualify:**

* **Calculate the z-score:**
    * z = (X - μ) / σ 
        * where:
            * X = cutoff score (80)
            * μ = mean (100)
            * σ = standard deviation (16)
    * z = (80 - 100) / 16 = -1.25

* **Find the area under the standard normal curve:**
    * Use a z-table or a calculator to find the area to the right of z = -1.25. 
        * This area represents the proportion of people with an IQ score of 80 or higher.

* **The area to the right of z = -1.25 is approximately 0.8944.**
* **Therefore, approximately 89.44% of people would qualify for admittance.**

**b. Minimum IQ score for 95% admittance:**

* **Find the z-score corresponding to the 95th percentile:**
    * Use a z-table or a calculator to find the z-score that corresponds to an area of 0.95 to the left of it. 
        * This z-score is approximately 1.645.

* **Calculate the corresponding IQ score:**
    * X = μ + z * σ
        * X = 100 + 1.645 * 16 
        * X = 126.32

* **Therefore, the minimum IQ score required for admittance should be approximately 126.**

**In summary:**

* **a.** 89.44% of people would qualify for admittance with an IQ cutoff of 80.
* **b.** To ensure 95% of children qualify, the minimum IQ score should be approximately 126.