The Ultimate Mean Median Mode Test! Quiz

The interquartile range (IQR) is a measure of statistical dispersion, specifically the range between the first quartile (Q1) and the third quartile (Q3) in a dataset. In this case, the IQR for the AGE attribute is 8. This means that the range between the age values at the 25th and 75th percentiles is 8 units. The IQR provides information about the spread or variability of the data, with a larger IQR indicating a greater dispersion of ages within the dataset.

When the largest value of a data set is doubled, it only affects the maximum value and does not impact the other values in the data set. Therefore, the interquartile range, which is the difference between the first quartile and the third quartile, remains the same. The mean, standard deviation, and range, on the other hand, are influenced by the change in the largest value and will increase.

The nominal level of measurement is represented by the variable "gender" because it is a categorical variable that represents different categories or groups. In this case, gender can be classified into two distinct categories: male and female. The nominal level of measurement does not involve any quantitative measurement or ranking, but rather focuses on classifying data into different categories or groups.

I. The median and the first quartile cannot be equal because the median is the middle value in a set of data, while the first quartile is the value below which 25% of the data falls. Since the median divides the data into two equal halves, it cannot be equal to the first quartile if the data is ordered from smallest to largest.

III. 2 is an outlier because it is significantly smaller than the other values in the set. An outlier is a value that is unusually small or large compared to the rest of the data.

Therefore, the correct answer is I and III only.

The proportion of students with a score of 33 or higher can be determined by finding the area under the normal distribution curve to the right of the value 33. Using the mean of 18 and standard deviation of 6, we can calculate the z-score for 33, which is (33-18)/6 = 2.5. By looking up the z-score of 2.5 in the z-table, we find that the proportion of students with a score of 33 or higher is approximately 0.0062.

The professor scaled the scores by multiplying the raw score by 1.2 and adding 15 points. This means that each student's score was increased by 20% (1.2 times) and then an additional 15 points were added. Therefore, the mean of the scaled scores will be 20% higher than the mean of the original scores, which is 51. So, the mean of the scaled scores is 51 + (0.20 * 51) = 76.2. The standard deviation of the scaled scores will remain the same since the scaling process does not affect the spread of the scores. Therefore, the standard deviation of the scaled scores is 5. Hence, the correct answer is 76.2 and 6.

The sample mean is calculated by adding up all the scores and dividing by the total number of scores. In this case, the sum of the scores is 417 and there are 20 scores. So, the sample mean is 417/20 = 20.85.

The standard deviation is a measure of how spread out the scores are from the mean. It is calculated by taking the square root of the average of the squared differences between each score and the mean. In this case, the standard deviation is 5.94.

The correct answer is that the scores of the males have a higher variability than the scores of the females. This can be inferred from the boxplots, where the box for the male scores is wider than the box for the female scores. This indicates that the male scores have a greater spread or range of values, suggesting higher variability.

The sum of the squared deviations of observations from the sample mean can be calculated using the formula: sum of (observation - mean)^2. Since the standard deviation is given as 3, we know that the mean deviation is also 3. Therefore, the sum of the squared deviations can be calculated as 20*(3^2) = 180. However, the question asks for the sum of the squared deviations, not the sum of the squared mean deviations. To convert from mean deviations to deviations, we need to multiply by the sample size, which is 20. Therefore, the correct answer is 180*20 = 3600, which is closest to 23.

Based on the given information, the child's IQ score of 113 is above the mean of 100 and therefore falls in the higher range of the distribution. Since the standard deviation is 15, it indicates that approximately 68% of the population falls within one standard deviation of the mean. As the child's IQ score is more than one standard deviation above the mean, it suggests that he is in the uppermost range of the distribution, which is represented by BAND -E.