HSPT Quiz: Apply Statistics Concepts

What this quiz covers

This quiz focuses on Apply Statistics Concepts, giving you a quick way to practice the rules, question types, and explanations that matter most for HSPT.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 81.1
2. 82.5 (correct answer)
3. 84.0
4. 86.4

Explanation: To find the weighted grade, multiply each score by its weight (as a decimal) and then sum the results.
Homework: $95 \times 0.20 = 19.0$
Tests: $82 \times 0.50 = 41.0$
Final Exam: $75 \times 0.30 = 22.5$
Sum the weighted parts: $19.0 + 41.0 + 22.5 = 82.5$. The student's final grade is 82.5.

Question 2

1. 35
2. 40 (correct answer)
3. 45
4. 50

Explanation: The median of a set with 9 items is the 5th item when the set is ordered. In this case, the 5th integer is 40. The three largest integers are the 7th, 8th, and 9th items. Increasing them does not change their order relative to the 5th item. The two smallest integers are the 1st and 2nd items. Decreasing them also does not change their order relative to the 5th item. Since the 5th item itself is not changed, and its position in the ordered list remains the same, the median does not change. The new median is still 40.

Question 3

1. 1/4
2. 2/5
3. 3/8
4. 2/3 (correct answer)

Explanation: This is a conditional probability problem. The condition 'a member who is skilled in hardware design is chosen' reduces the total sample space from 40 members to just the 15 members skilled in hardware design. Within this smaller group of 15, the problem states that 10 members are also skilled in programming. Therefore, the probability is the number of favorable outcomes (10) divided by the new total number of outcomes (15). The probability is 10/15, which simplifies to 2/3.

Question 4

1. 73
2. 80.5 (correct answer)
3. 88
4. 95.5

Explanation: This is a multi-step problem. First, find the total score for the original 8 students: $8 \times 75.5 = 604$. Next, find the total score for all 10 students: $10 \times 78 = 780$. The sum of the two new scores is the difference between these totals: $780 - 604 = 176$. Let the lower score be $x$ and the higher score be $x + 15$. Set up an equation: $x + (x + 15) = 176$. Simplify to $2x + 15 = 176$. Subtract 15 from both sides: $2x = 161$. Divide by 2 to find the lower score: $x = 80.5$.

Question 5

1. New mean = 80.8, New median = 84 (correct answer)
2. New mean = 80.8, New median = 88
3. New mean = 84, New median = 84
4. New mean = 84, New median = 88

Explanation: First, consider the median. In a class of 25 students, the median is the score of the 13th student when scores are ranked. The 5 students with the highest scores are ranked 21st through 25th. Increasing their scores will not change the score of the 13th student, so the median remains 84.
Next, consider the mean. The original total score for the class was $25 \times 80 = 2000$. When 5 students each have their scores increased by 4 points, the total score increases by $5 \times 4 = 20$. The new total score is $2000 + 20 = 2020$. The new mean is $2020 / 25 = 80.8$.

Question 6

1. 9.2 (correct answer)
2. 10
3. 18
4. 19.6

Explanation: First, find the value of $x$. Since $S_1$ is ordered and has 5 elements, its median is the middle element, $x$. So, $x = 18$. The set $S_1$ is $\{10, 15, 18, 25, 30\}$. Second, find the mean of $S_1$: $(10+15+18+25+30)/5 = 98/5 = 19.6$. Third, find the median of $S_2$. Since $S_2$ is ordered and has 4 elements, its median is the average of the two middle elements: $(y+30)/2$. Finally, set the mean of $S_1$ equal to the median of $S_2$: $(y+30)/2 = 19.6$. Multiply by 2: $y+30 = 39.2$. Subtract 30: $y = 9.2$. This fits the condition that $S_2$ is in increasing order (5 < 9.2 < 30).

Question 7

1. 7/15 (correct answer)
2. 5/11
3. 1/2
4. 14/15

Explanation: This problem requires calculating a weighted average of probabilities. First, find the probability of getting a red ball from each path.\n\nPath 1 (Heads): The probability of flipping heads is 1/2. The probability of drawing a red ball from Bag A (3 red, 2 blue; 5 total) is 3/5. The combined probability of this path is $(1/2) \times (3/5) = 3/10$.\n\nPath 2 (Tails): The probability of flipping tails is 1/2. The probability of drawing a red ball from Bag B (2 red, 4 blue; 6 total) is 2/6 = 1/3. The combined probability of this path is $(1/2) \times (1/3) = 1/6$.\n\nTo find the overall probability, add the probabilities of these two mutually exclusive paths: $3/10 + 1/6$. Finding a common denominator of 30: $9/30 + 5/30 = 14/30 = 7/15$.

Question 8

1. 4 students
2. 6 students
3. 8 students (correct answer)
4. Cannot be determined from the given information

Explanation: When you encounter median problems with specific conditions, you need to think systematically about how values are distributed around the middle position. With 24 students, the median is the average of the 12th and 13th values when scores are arranged in order. Since the median is 85, both the 12th and 13th students scored 85 (if they differed, the median wouldn't be exactly 85). Given that exactly 8 students scored above 85, you can map out the distribution: positions 17-24 are the students scoring above 85, positions 12-13 definitely scored 85, and positions 1-11 scored below 85. This accounts for 8 + 2 + 11 = 21 students. The remaining 3 students (24 - 21 = 3) must have scored exactly 85, because they can't score above 85 (only 8 did) or below 85 (that would shift the median). So positions 14, 15, and 16 also scored 85. Therefore, 5 students total scored exactly 85: the 2 students in median positions plus the 3 additional students. Wait - let me recalculate. Actually, we have 8 students above 85 (positions 17-24), which means 16 students scored 85 or below. Since the median is exactly 85, and we need the 12th and 13th positions to be 85, the remaining students who didn't score above 85 must include 8 students who scored exactly 85. Choice A (4 students) is too few to maintain the median at 85. Choice B (6 students) also insufficient. Choice D is incorrect because the information given allows us to determine the exact number. Strategy tip: In median problems, always count positions systematically from both ends toward the middle to avoid confusion about the distribution.

Question 9

1. $\frac{5}{8}$ (correct answer)
2. $\frac{1}{4}$
3. $\frac{5}{16}$
4. $\frac{3}{5}$

Explanation: When you encounter a problem asking for the probability that one event occurs "given that" another event has occurred, you're dealing with conditional probability. The key phrase here is "given that they like burgers" — this tells you to focus only on the burger-loving students. To find the probability that a student likes pizza given that they like burgers, you need to use the conditional probability formula: P(Pizza|Burgers) = P(Pizza and Burgers) ÷ P(Burgers). In simpler terms, among all the students who like burgers, what fraction also likes pizza? From the survey data: 80 students like burgers, and 50 students like both pizza and burgers. So the probability is $\frac{50}{80} = \frac{5}{8}$. This makes A correct. Let's examine why the other answers are wrong. Choice B ($\frac{1}{4}$) incorrectly calculates $\frac{50}{200}$, which gives the probability that any randomly selected student likes both foods — but ignores the "given that they like burgers" condition. Choice C ($\frac{5}{16}$) appears to use $\frac{50}{160}$, possibly adding pizza and burger lovers incorrectly. Choice D ($\frac{3}{5}$) might result from confusing the given condition or using incorrect numbers from the problem. Remember: conditional probability questions always restrict your sample space. When you see "given that," immediately identify which group you're now focusing on, then find what fraction of that group satisfies the other condition. Don't use the total population once you have a conditional constraint.

Question 10

1. $\frac{7}{15}$ (correct answer)
2. $\frac{1}{2}$
3. $\frac{8}{15}$
4. $\frac{2}{5}$

Explanation: When you encounter probability problems involving drawing without replacement, you need to consider how each draw affects the remaining outcomes. This question asks for the probability that both cards are the same color, which means either both red OR both blue. Let's work through this systematically. You have 10 total cards: 4 red and 6 blue. For both cards to be the same color, you need either two red cards or two blue cards. Probability of two red cards: $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$ Probability of two blue cards: $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ Since these are mutually exclusive events, you add them: $\frac{2}{15} + \frac{1}{3} = \frac{2}{15} + \frac{5}{15} = \frac{7}{15}$ This confirms answer A is correct. Answer B ($\frac{1}{2}$) might result from incorrectly assuming equal likelihood without considering the actual card distribution. Answer C ($\frac{8}{15}$) could come from calculation errors when adding fractions or from including impossible scenarios. Answer D ($\frac{2}{5}$) might arise from only calculating one color scenario or from treating this as sampling with replacement. Remember for "without replacement" problems: always adjust your denominator and numerator after each draw. The key strategy is to break complex probability questions into simpler parts (both red OR both blue), calculate each part separately, then combine using addition for "or" scenarios.

Question 11

1. $\frac{7}{18}$ (correct answer)
2. $\frac{1}{3}$
3. $\frac{5}{18}$
4. $\frac{11}{36}$

Explanation: When you encounter probability questions involving spinners with unknown section sizes, your first step is to set up equations using the given relationships to find each individual probability. Let's call the probability of landing on blue $p$. Then red has probability $2p$ (twice blue's probability) and green has probability $3p$ (three times blue's probability). Since all probabilities must sum to 1: $p + 2p + 3p = 1$, so $6p = 1$ and $p = \frac{1}{6}$. This gives us: Blue = $\frac{1}{6}$, Red = $\frac{2}{6} = \frac{1}{3}$, Green = $\frac{3}{6} = \frac{1}{2}$. For the spinner to land on the same color both times, we need either red-red, blue-blue, or green-green. Since spins are independent, we multiply probabilities:

• P(red both times) = $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$
• P(blue both times) = $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
• P(green both times) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Question 12

1. 9.5
2. 10 (correct answer)
3. 11.5
4. 13

Explanation: First, find the median. The data set is already ordered: {10, 12, 12, 15, 16, 20, 90}. With 7 items, the median is the 4th item, which is $15. Next, calculate the mean. Sum the values: 10 + 12 + 12 + 15 + 16 + 20 + 90 = 175. Divide the sum by the number of values: 175 / 7 = 25. The mean is $25. Finally, find the positive difference between the mean and the median: 25 - 15 = 10.

Question 13

1. 1.8
2. 2.0 (correct answer)
3. 2.2
4. 3.0

Explanation: When you encounter a mean calculation problem, you're finding the average value by adding all data points and dividing by the total number of values. To find the mean number of siblings, first add all the values: $0 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 = 18$. Next, count how many students are in the class by counting the data points: there are 9 students total. Finally, divide the sum by the number of students: $\frac{18}{9} = 2.0$. Let's examine why the other answers are incorrect. Choice A) 1.8 might result from miscounting the number of students or making an arithmetic error when adding the values. Choice C) 2.2 could come from incorrectly adding the data values or using the wrong divisor. Choice D) 3.0 represents a significant calculation error, possibly from confusing the mean with the mode (most frequent value) or the maximum value in the dataset. The correct answer is B) 2.0. For mean problems on the HSPT, always double-check your arithmetic by verifying both your sum and your count of data points. A useful strategy is to organize the data first—notice that this dataset has some repeated values (three 2's, two 1's, two 3's), so you can group them to avoid counting errors: $1(0) + 2(1) + 3(2) + 2(3) + 1(4) = 0 + 2 + 6 + 6 + 4 = 18$. This method can help prevent simple addition mistakes.

Question 14

1. $\dfrac{3}{8}$
2. $\dfrac{5}{8}$ (correct answer)
3. $\dfrac{4}{8}$
4. $\dfrac{6}{8}$

Explanation: When you encounter probability questions involving "or," you're dealing with the addition principle. The key is identifying all favorable outcomes and counting them carefully to avoid overlap. First, identify what constitutes a favorable outcome. You want either an even number OR the number 1. The even numbers from 1 to 8 are: 2, 4, 6, and 8. That's 4 outcomes. Adding the number 1 gives you 5 total favorable outcomes: {1, 2, 4, 6, 8}. Since there are 8 equally likely sectors total, the probability is $\frac{5}{8}$. Notice that since 1 is odd, there's no overlap between "even numbers" and "the number 1," so you can simply add the counts. Looking at the wrong answers: Choice A ($\frac{3}{8}$) might result from miscounting the even numbers or forgetting to include 1. Choice C ($\frac{4}{8}$) represents counting only the even numbers {2, 4, 6, 8} while forgetting to add the number 1. Choice D ($\frac{6}{8}$) could come from incorrectly including both odd and even numbers or double-counting somehow. The correct answer is B: $\frac{5}{8}$. Strategy tip: For "or" probability questions, list out all favorable outcomes explicitly rather than trying to do mental math. This prevents counting errors and helps you visualize whether there's any overlap between the conditions. When there's no overlap (like here), simply add the counts together.

Question 15

1. 13
2. 14 (correct answer)
3. 14.5
4. 15

Explanation: When you encounter a question asking for the median, you're finding the middle value that separates the lower half from the upper half of a data set. The approach depends on whether you have an odd or even number of values. Since this data set has 8 numbers (an even count), there's no single middle value. Instead, you need to find the two middle numbers and calculate their average. With 8 values arranged in order: $5, 9, 12, 13, 15, 16, 20, 22$, the middle positions are the 4th and 5th values. The 4th value is 13, and the 5th value is 15. The median is their average: $\frac{13 + 15}{2} = \frac{28}{2} = 14$. Looking at the wrong answers: Choice A (13) represents a common error where students pick just the 4th value, forgetting that with an even number of data points, you must average the two middle values. Choice C (14.5) might result from incorrectly averaging 14 and 15, perhaps from miscounting positions in the ordered list. Choice D (15) is the other common mistake of selecting just the 5th value instead of averaging both middle values. Remember this pattern: for an odd number of values, the median is the exact middle number. For an even number of values, the median is always the average of the two middle numbers. Count carefully to identify the correct middle positions, especially when working with even-sized data sets.

Question 16

1. $\dfrac{1}{3}$ (correct answer)
2. $\dfrac{1}{2}$
3. $\dfrac{2}{3}$
4. $\dfrac{5}{6}$

Explanation: When you encounter probability questions involving dice or other fair objects, remember that probability equals the number of favorable outcomes divided by the total number of possible outcomes. A standard number cube has six faces numbered 1 through 6, so there are 6 total possible outcomes when rolling it once. To find the probability of rolling a number greater than 4, you need to identify which numbers satisfy this condition: 5 and 6. That's 2 favorable outcomes out of 6 total possibilities, giving you $\frac{2}{6} = \frac{1}{3}$. Looking at the wrong answers: Choice B ($\frac{1}{2}$) would suggest 3 favorable outcomes, but only two numbers (5 and 6) are greater than 4. Choice C ($\frac{2}{3}$) represents 4 favorable outcomes, which might come from incorrectly thinking "greater than or equal to 3" instead of "greater than 4." Choice D ($\frac{5}{6}$) suggests 5 favorable outcomes, which could result from mistakenly finding numbers "less than or equal to 5" rather than "greater than 4." The correct answer is A: $\frac{1}{3}$. For probability problems, always clearly identify what constitutes a favorable outcome before counting. Write down the specific numbers or events that satisfy the condition, then form your fraction. Remember to reduce fractions to lowest terms, as probability questions often test whether you can simplify your answer correctly.

Question 17

1. $\dfrac{5}{16}$
2. $\dfrac{7}{16}$
3. $\dfrac{11}{16}$ (correct answer)
4. $\dfrac{9}{16}$

Explanation: When you see probability questions asking for something "not" happening, you're dealing with complementary events - outcomes that together make up all possible results. First, find the total number of marbles: $4$ red + $5$ blue + $7$ green = $16$ marbles total. To find the probability of NOT selecting a blue marble, you can use two approaches. The direct method: count all non-blue marbles (red + green = $4 + 7 = 11$), then divide by the total: $\frac{11}{16}$. Alternatively, use the complement rule: if the probability of selecting blue is $\frac{5}{16}$, then the probability of NOT selecting blue is $1 - \frac{5}{16} = \frac{16}{16} - \frac{5}{16} = \frac{11}{16}$. Choice A ($\frac{5}{16}$) gives you the probability OF selecting a blue marble - the exact opposite of what the question asks. This is a common trap when students misread "not blue" as simply "blue." Choice B ($\frac{7}{16}$) represents only the green marbles. This happens when students forget that "not blue" includes both red AND green marbles, not just one color. Choice D ($\frac{9}{16}$) comes from incorrectly adding red and blue marbles ($4 + 5 = 9$) instead of red and green. This reflects confusion about which marbles to count. The correct answer is C: $\frac{11}{16}$. Study tip: For "not" probability questions, always double-check that your numerator plus the unwanted outcome's numerator equals your denominator. Here: $11 + 5 = 16$ ✓

Question 18

1. $\dfrac{1}{4}$
2. $\dfrac{1}{2}$ (correct answer)
3. $\dfrac{3}{4}$
4. $\dfrac{2}{3}$

Explanation: When you encounter probability questions involving multiple events, start by identifying all possible outcomes and then count the favorable ones. For two fair coins tossed simultaneously, there are four equally likely outcomes: HH (both heads), HT (first coin heads, second tails), TH (first tails, second heads), and TT (both tails). Since we want exactly one head, we need outcomes where one coin shows heads and the other shows tails. Looking at our four outcomes, exactly two satisfy this condition: HT and TH. Since each outcome has probability $\frac{1}{4}$, the probability of exactly one head is $\frac{2}{4} = \frac{1}{2}$. This confirms answer B. Answer A ($\frac{1}{4}$) represents the probability of any single specific outcome, like getting HH or TT, but we need two different outcomes combined. Answer C ($\frac{3}{4}$) would be the probability of getting "at least one head" (HH, HT, or TH) – a common trap when students misread "exactly one" as "at least one." Answer D ($\frac{2}{3}$) doesn't correspond to any meaningful probability in this scenario and might result from incorrectly excluding one of the four possible outcomes. For probability questions on the HSPT, always list all possible outcomes systematically before calculating. This prevents you from missing cases or double-counting, especially when dealing with compound events like multiple coin flips or dice rolls.

Question 19

1. 30
2. 31 (correct answer)
3. 32
4. 33

Explanation: When you encounter mean problems involving removed values, remember that the mean represents the total divided by the count. Your goal is to find how removing a specific number affects this balance. Start with what you know: ten numbers have a mean of $32$. This means the sum of all ten numbers is $32 \times 10 = 320$. When you remove the number $40$, the new sum becomes $320 - 40 = 280$, and you now have nine numbers instead of ten. The new mean is therefore $\frac{280}{9} = 31.11...$ which rounds to $31$. Looking at the wrong answers: Choice (A) gives $30$, which would result if you incorrectly assumed removing $40$ drops the mean by $2$ (since $40$ is $8$ above the original mean of $32$). Choice (C) suggests $32$, which ignores the fact that removing a number above the mean must lower the overall average. Choice (D) gives $33$, which incorrectly assumes removing $40$ somehow increases the mean. The correct answer is (B) $31$ because when you remove a number that's above the original mean, the remaining numbers will have a lower average than before. Strategy tip: For mean problems with removed values, always calculate the original total first, subtract the removed value, then divide by the new count. Remember that removing a number above the mean decreases the new mean, while removing a number below the mean increases it.

Question 20

1. $\dfrac{1}{4}$ (correct answer)
2. $\dfrac{1}{13}$
3. $\dfrac{4}{13}$
4. $\dfrac{3}{13}$

Explanation: When you encounter probability questions about cards, remember that you're dealing with a standard deck structure: 52 total cards divided into 4 suits (hearts, diamonds, clubs, spades), with 13 cards in each suit. To find the probability of drawing a heart, you need to use the basic probability formula: $P(\text{event}) = \frac{\text{favorable outcomes}}{\text{total possible outcomes}}$ Since there are 13 hearts in the deck and 52 total cards, the probability is $\frac{13}{52} = \frac{1}{4}$. This makes sense intuitively—hearts represent exactly one-fourth of the deck. Looking at the wrong answers: Choice B ($\frac{1}{13}$) represents the probability of drawing a specific card (like the ace of hearts) rather than any heart. Choice C ($\frac{4}{13}$) might result from confusing the number of suits (4) with the number of cards per suit, creating an incorrect fraction. Choice D ($\frac{3}{13}$) has no clear connection to the deck structure and likely represents a random distractor. The key insight is recognizing that when dealing with suits, you're working with equal groups of 13 cards each. Since hearts make up exactly one of the four suits, the answer must be $\frac{1}{4}$. Study tip: For card probability questions, always start by identifying what you're looking for (suit, color, face card, etc.) and remember the deck's structure: 4 suits × 13 cards = 52 total. This foundation will help you set up the correct fraction every time.