Question

in one math class, the first exam had an average score of 63.1% with a standard deviation of 11.2%. the second exam had a mean score of 61.3%, with a standard deviation of 20.9%.
based on these numbers, tell whether each statement is true or false.
part 1 of 6
(a) there were probably some students that took a zero on exam 1.
the statement is
correct answer
the statement is false.
part 2 of 6
(b) even though the means were similar, the spread of the scores was wider on the second exam.
the statement is
part: 2 / 6
part 3 of 6
(c) most students did better on the second exam.
the statement is

Explanation:

Step1: Analyze the possibility of a zero on Exam I

The mean of Exam I is \(63.1\%\) and the standard deviation is \(11.2\%\). Using the empirical rule (for a normal distribution, about \(99.7\%\) of data lies within \(3\) standard deviations of the mean). The lower bound within \(3\) standard deviations is \(63.1-3\times11.2 = 63.1 - 33.6=29.5\%\). A score of \(0\%\) is more than \(3\) standard deviations below the mean. While it's not impossible (as distributions may not be perfectly normal), it's very unlikely.

Step2: Analyze the spread of scores

The spread of data is measured by the standard deviation. For Exam I, \(\sigma_1 = 11.2\%\) and for Exam II, \(\sigma_2=20.9\%\). Since \(20.9>11.2\), the spread (variability) of scores is wider on the second exam.

Step3: Analyze if most students did better on the second exam

The mean of Exam I (\(63.1\%\)) is higher than the mean of Exam II (\(61.3\%\)). The mean is a measure of central tendency. A higher mean for Exam I suggests that, on average, students scored higher on Exam I. There is no information about the individual - level performance (e.g., median, mode, or individual data points) to support that most students did better on the second exam.

Answer:

(a) The statement is false.
(b) The statement is true.
(c) The statement is false.