Final answer:
The probability of selecting a male student is 0.25. The proportion between z = -0.5 and z = 0.5 on a normal distribution is about 76.58%, and the percentage above a z-value of 1.53 is approximately 6.3%. Susan's z-score on the biology exam is 2, indicating her performance was well above average.
Step-by-step explanation:
Understanding Probability and the Standard Normal Distribution
To find the probability of selecting a male from a class of 10 males and 30 females, we simply divide the number of males by the total number of students. The probability is 10/(10+30) = 10/40 = 1/4 or 0.25.
For the proportion of scores between z = -0.5 and z = 0.5 on a normal distribution, we use a z-table or a standard normal distribution table. The area under the curve between these two z-values represents the probability. Typically, it is around 38.29% for each side, summing up to about 76.58% for the total area between these two z-scores.
To find the area above a z-value of 1.53, we also refer to the z-table. The table provides the area to the left of the z-score, so we subtract that value from the total area under the curve, which is 1 (or 100%). If the z-table shows 0.937 for z = 1.53, the area to the right is 1 - 0.937 = 0.063 or 6.3%.
Z-scores are calculated by taking the raw score, subtracting the mean, and dividing by the standard deviation. Susan's exam score of 95, with a mean of 85 and a standard deviation of 5, gives her a z-score of (95-85)/5 = 2. This means Susan's score was 2 standard deviations above the mean. Interpretation of this score indicates that Susan performed significantly above average on her biology final exam.