A university administrator expects that 25% of students in a core course will receive an A. He looks at the grades assigned to 60 students. The probability that the proportion of students who receive an A is not between 0.20 and 0.30 is ________.
A) 0.1867
B) 0.3711
C) 0.6266
D) 0.8133

Question

Quizplus · Accepted Answer

This problem can be solved using the normal approximation to the binomial distribution. The expected proportion of students receiving an A is 0.25, and we are interested in the probability that the observed proportion is not between 0.20 and 0.30 in a sample of 60 students. First, calculate the mean ($\mu$) and standard deviation ($\sigma$) of the binomial distribution:- $\mu = np = 60 \times 0.25 = 15$- $\sigma = \sqrt{np(1-p)} = \sqrt{60 \times 0.25 \times (1-0.25)} = \sqrt{11.25} \approx 3.354$To use the normal approximation, convert the proportions 0.20 and 0.30 to counts:- Lower bound: 0.20 of 60 students = 12- Upper bound: 0.30 of 60 students = 18Next, convert these counts to z-scores:- $Z_{lower} = \frac{12 - 15}{3.354} \approx -0.894$- $Z_{upper} = \frac{18 - 15}{3.354} \approx 0.894$Using standard normal distribution tables or a calculator, find the probabilities corresponding to these z-scores. The probability of being between these z-scores is approximately 0.3711. However, since we are interested in the probability of NOT being between these scores, we subtract this from 1:- $P(\text{not between 0.20 and 0.30}) = 1 - P(\text{between 0.20 and 0.30}) = 1 - 0.3711 = 0.6289$The answer closest to this calculated probability is 0.3711, which corresponds to the probability of being between 0.20 and 0.30, indicating a misunderstanding in the final step. The correct interpretation should focus on the calculation method and the closest provided option, acknowledging the approximation and potential rounding errors in the calculation process.

A University Administrator Expects That 25% of Students in a Core