ISAT - International Student Admissions Probability and Statistical Analysis Questions and Answers — Questions and Answers
Question 1: A research group is studying the weekly screen time of teenagers. They collect the following data in hours from a sample of 7 participants: 15, 18, 20, 22, 25, 28, 75. Which measure of central tendency would most accurately describe a typical teenager's screen time in this sample?
- The mean, because it includes all data points.
- The mode, because it represents the most frequent value.
- The median, because it is not significantly affected by the outlier. (Correct answer)
- The range, because it shows the full spread of the data.
Correct answer: The median, because it is not significantly affected by the outlier.
The dataset contains an outlier (75 hours), which is significantly higher than the other values. The mean would be skewed upward by this outlier (to approx. 29 hours), not reflecting the typical value. The median, which is the middle value (22 hours), is resistant to such outliers and therefore provides a more accurate representation of the central tendency for this skewed dataset.
Question 2: A city's health department observes that during summer months, sales of ice cream are strongly and positively correlated with the number of heatstroke incidents. They conclude that consuming ice cream causes heatstroke. Which of the following statements best describes the flaw in their conclusion?
- The conclusion is valid as the statistical correlation is strong.
- The sample size was likely too small to draw a valid conclusion.
- The direction of causality is reversed; heatstroke causes people to eat ice cream.
- A third variable, high temperatures, is likely the cause of both increased ice cream sales and more heatstroke incidents. (Correct answer)
Correct answer: A third variable, high temperatures, is likely the cause of both increased ice cream sales and more heatstroke incidents.
This is a classic example of the logical fallacy 'correlation does not imply causation'. While ice cream sales and heatstroke incidents are correlated, one does not cause the other. A confounding or third variable, in this case, hot weather, is the actual cause for both phenomena to increase simultaneously.
Question 3: A quality control inspector is examining two production lines, Line A and Line B, which both produce bolts with a target length of 50 mm. The inspector takes a sample from each line. Line A's sample has lengths (in mm) of {49, 51, 50, 49, 51}. Line B's sample has lengths of {45, 55, 50, 47, 53}. Both samples have a mean of 50 mm. Which of the following statements is true regarding the samples?
- Line A has a larger standard deviation than Line B.
- Line B has a larger standard deviation than Line A. (Correct answer)
- Both lines have the same standard deviation.
- The standard deviation cannot be determined without more data.
Correct answer: Line B has a larger standard deviation than Line A.
Standard deviation measures the dispersion or spread of data points around the mean. The bolt lengths from Line B ({45, 55, 50, 47, 53}) are more spread out from the mean of 50 mm compared to the lengths from Line A ({49, 51, 50, 49, 51}), which are tightly clustered around the mean. Therefore, Line B's sample has a larger standard deviation, indicating less consistency.
Question 4: An office has 100 employees. 60% of the employees are women. 30% of all employees have advanced degrees. Of the employees with advanced degrees, 20 are women. What is the probability that a randomly selected employee is a man who does not have an advanced degree?
- 0.20
- 0.30 (Correct answer)
- 0.40
- 0.10
Correct answer: 0.30
Total employees = 100. Women = 60, Men = 40. Employees with advanced degrees = 30. Women with advanced degrees = 20. Therefore, Men with advanced degrees = 30 - 20 = 10. The total number of men is 40. So, the number of men who do not have an advanced degree is 40 (total men) - 10 (men with advanced degrees) = 30. The probability of selecting a man without an advanced degree is 30/100 = 0.30.
Question 5: In a class of 50 students, 30 students are enrolled in Chemistry and 25 students are enrolled in Physics. 10 students are enrolled in both subjects. What is the probability that a randomly selected student is enrolled in Chemistry, given that they are enrolled in Physics?
- 1/3
- 2/5 (Correct answer)
- 1/5
- 3/5
Correct answer: 2/5
This is a conditional probability problem. The formula is P(A|B) = P(A and B) / P(B). Let A be the event that a student is in Chemistry, and B be the event that a student is in Physics. We are looking for P(Chemistry | Physics). The number of students in both is 10. The total number of students in Physics is 25. Therefore, the probability that a student is in Chemistry given they are in Physics is the number of students in both subjects divided by the total number of students in Physics: 10/25 = 2/5.
Question 6: A box contains 6 red balls, 4 blue balls, and 5 green balls. If two balls are drawn from the box one after the other without replacement, what is the probability that both balls are red?
- 1/7 (Correct answer)
- 1/5
- 4/25
- 1/9
Correct answer: 1/7
The probability of the first event (drawing a red ball) is 6/15. Since the first ball is not replaced, there are now only 14 balls left in the box, and only 5 of them are red. The probability of the second event (drawing another red ball) is 5/14. To find the probability of both events happening, we multiply their probabilities: (6/15) * (5/14) = 30/210, which simplifies to 1/7.
ISAT - International Student Admissions Probability and Statistical Analysis Questions and Answers
A research group is studying the weekly screen time of teenagers.
They collect the following data in hours from a sample of 7 participants: 15, 18, 20, 22, 25, 28, 75.
Which measure of central tendency would most accurately describe a typical teenager's screen time in this sample?