1. **State the problem:** We have 600 applicants with IQs approximately normally distributed with mean $\mu=115$ and standard deviation $\sigma=12$. The college requires an IQ of at least 95. We want to find how many applicants have IQ less than 95 and thus will be rejected. 2. **Standardize the cutoff IQ:** Convert the IQ cutoff 95 to a standard normal variable $Z$ using $$Z=\frac{X-\mu}{\sigma}=\frac{95-115}{12} = \frac{-20}{12} = -1.6667.$$ 3. **Find the probability of rejection:** We want $P(X<95) = P(Z < -1.6667)$. Using standard normal distribution tables or a calculator, $$P(Z < -1.6667) \approx 0.0478.$$ 4. **Calculate the number of rejected applicants:** Multiply this probability by the total number of applicants: $$600 \times 0.0478 = 28.68.$$ Since the number of applicants must be an integer, approximately 29 applicants will be rejected based on IQ. **Final answer:** Approximately **29** applicants will be rejected due to having IQ less than 95.