Question

some iq tests are standardized to a normal model n(100,17). a) what cutoff value bounds the highest 15% of all iqs? b) what cutoff value bounds the lowest 25% of the iqs? c) what cutoff values bound the middle 80% of the iqs?

Explanation:

Step1: Recall z - score formula

For a normal distribution $N(\mu,\sigma)$ with mean $\mu$ and standard - deviation $\sigma$, the z - score is given by $z=\frac{x - \mu}{\sigma}$, and we can find the z - scores corresponding to the given percentiles using the standard normal distribution table (z - table). Here, $\mu = 100$ and $\sigma=17$.

Step2: Find z - score for part (a)

The highest 15% means the area to the right of the z - score is 0.15, so the area to the left is $1 - 0.15=0.85$. Looking up in the z - table, the z - score $z_{1}$ corresponding to an area of 0.85 is approximately 1.04. Then, using the z - score formula $z=\frac{x - \mu}{\sigma}$, we have $1.04=\frac{x - 100}{17}$. Solving for $x$ gives $x = 100+1.04\times17=100 + 17.68 = 117.68$.

Step3: Find z - score for part (b)

The lowest 25% means the area to the left of the z - score is 0.25. Looking up in the z - table, the z - score $z_{2}$ corresponding to an area of 0.25 is approximately - 0.67. Using the z - score formula $z=\frac{x - \mu}{\sigma}$, we have $-0.67=\frac{x - 100}{17}$. Solving for $x$ gives $x=100-0.67\times17=100 - 11.39 = 88.61$.

Step4: Find z - scores for part (c)

The middle 80% means the area in the two tails is $1 - 0.80 = 0.20$, so the area in each tail is $\frac{0.20}{2}=0.10$. The z - score corresponding to the left - hand tail (area of 0.10) is $z_{3}\approx - 1.28$ and the z - score corresponding to the right - hand tail (area of $1 - 0.10 = 0.90$) is $z_{4}\approx1.28$. For $z=-1.28$, $x_{1}=100-1.28\times17=100 - 21.76 = 78.24$. For $z = 1.28$, $x_{2}=100+1.28\times17=100 + 21.76 = 121.76$.

Answer:

a. 117.68
b. 88.61
c. 78.24 and 121.76