Develop Non-Uniform Probability Models - 7th Grade Math

Use the table to answer the question. A coin was flipped 100 times with the following results by 10-flip intervals. Based on this data, how should Rosa revise her probability model for getting heads, and what does the pattern suggest about the coin?
Answer: $P(Heads) = 0.64$ ; the coin appears biased toward heads since 64 > 50 and the trend is increasing Based on observed data, $P(Heads) = \frac{64}{100} = 0.64$ . Since this exceeds 0.50 and shows an increasing trend in later intervals, the coin appears biased toward heads. Choice B incorrectly maintains the theoretical probability despite evidence. Choice C reaches the right probability but overstates the certainty of bias. Choice D confuses heads and tails probabilities.
A class uses a weighted spinner with three colors. In 100 practice spins, the results were: Red 45, Blue 30, Green 25. Which probability model best matches the practice data, and does it represent a uniform or non-uniform model?
Answer: P(R)=0.45, P(B)=0.30, P(G)=0.25; non-uniform Dividing each color's spins by the total gives the relative frequencies: 45/100 = 0.45 for red, 30/100 = 0.30 for blue, and 25/100 = 0.25 for green, which sum to 1 and match choice C. Since the three probabilities aren't equal, this is a non-uniform model, not uniform. Choice B swaps the blue and green probabilities and incorrectly calls the model uniform. Choice A uses the raw counts (45, 30, 25) instead of converting them to probabilities, and choice D assumes all three colors are equally likely, which the data doesn't support.
A music app tracked what a student listened to on 60 days: Pop 27 days, Rap 21 days, Other 12 days. Using a probability model based on these frequencies, what is $P(\text{Pop})$ ?
Answer: $\frac{27}{60}=0.45$ To build a probability model from frequencies, divide the number of days for the outcome by the total number of days: P(Pop)=27/60=0.45. Choice B assumes all three categories are equally likely, which ignores the actual data. Choice C uses the frequency for Rap (21) instead of Pop, mixing up the categories. Choice D inverts the fraction, which does not represent a valid probability.
A spinner has four sections labeled A, B, C, and D. After 200 spins, the results were: A 70, B 50, C 60, D 20. Which probability model best matches the results?
Answer: $P(A)=\frac{70}{200}=0.35,\;P(B)=\frac{50}{200}=0.25,\;P(C)=\frac{60}{200}=0.30,\;P(D)=\frac{20}{200}=0.10$ This question tests developing a non-uniform model from spinner frequencies over 200 spins, using relative frequencies. Non-uniform: unequal; develop by observing A70, B50, C60, D20, calculating 70/200=0.35, etc., sum=1. For example, P(A)=0.35, P(B)=0.25, P(C)=0.30, P(D)=0.10, matching data. The correct model is B, with accurate relative frequencies. Errors: dividing by 100 in A, uniform in C, swapping in D. Developing: (1) collect, (2) calculate relative, (3) assign, (4) verify sum=1. Non-uniform from sections; mistakes: wrong denominator or assuming equal.
A music app tracked what genre a student listened to over 60 songs: Pop 27, Hip-hop 15, Rock 18. Using a probability model based on these frequencies, what is $P(\text{Pop})$ ?
Answer: $\frac{27}{60}=0.45$ Since the student listened to 60 total songs and 27 of them were Pop, the probability model gives P(Pop) as 27 divided by 60, which equals 0.45, matching choice C. Choice A, 27/100, incorrectly divides by 100 instead of the actual total number of songs, 60. Choice B, 1/3, assumes each genre is equally likely, which ignores the actual, uneven listening data. Choice D, 60/27, flips the fraction, dividing the total number of songs by the Pop count instead of the other way around. Building a probability model from frequency data always means dividing each category's count by the true total number of observations, which here is 60 songs, not 100.
A class uses a weighted spinner with 3 colors. In 100 practice spins, the results were: Red 45, Blue 30, Green 25. Which probability model best matches the data?
Answer: P(red)=0.45, P(blue)=0.30, P(green)=0.25 To build a probability model from the data, divide each color's count by the total number of spins: red is 45/100=0.45, blue is 30/100=0.30, and green is 25/100=0.25. These three probabilities add up to 1.00, confirming they form a valid model, matching Choice D. Choice A uses different numbers that do not match the actual data from the 100 spins. Choice B assumes each color is equally likely, which ignores the fact that the spinner is weighted and the data shows unequal results. Choice C uses the raw counts, 45, 30, and 25, directly as probabilities, but these numbers add up to 100, not 1, so they cannot be valid probabilities.
A weather station tracks rainy days over 200 days and finds it rained on 45 days. However, they notice that in the first 100 days, it rained 15 times, while in the second 100 days, it rained 30 times. How should they develop their probability model for rain, and what does this pattern indicate?
Answer: Use $P(rain) = \frac{45}{200} = 0.225$ for the entire period, recognizing this may mask seasonal variation in rainfall The overall probability $P(rain) = \frac{45}{200} = 0.225$ uses all available data, but the significant difference between periods (0.15 vs 0.30) suggests seasonal variation that a single probability model may not capture well. Choice B arbitrarily favors recent data. Choice C arbitrarily favors early data. Choice D incorrectly averages probabilities rather than using frequency data directly.
A spinner is tested 100 times with these results: Red 45, Blue 30, Green 25. Using the probability model from these results, how many red results would you predict in 50 more spins?
Answer: 22.5 This question tests using a non-uniform probability model developed from observed spinner frequencies to predict outcomes in additional spins. Non-uniform means colors have different probabilities, developed from 100 spins (red 45, blue 30, green 25) by calculating P(red)=0.45, then predicting for 50 spins as 50×0.45=22.5 expected reds. For example, with this model, the prediction is 22.5, which is a theoretical average, not requiring a whole number. The correct prediction uses the relative frequency multiplied by the number of trials. Common mistakes include using raw counts or other probabilities. To use the model: multiply probability by number of trials for expected value. Comparing predictions to observations helps assess model fit, and non-uniform models account for biases in the spinner.
A student creates a probability model from 100 trials of a spinner: $P(\text{red})=0.45$ , $P(\text{blue})=0.30$ , $P(\text{green})=0.25$ . Which statement correctly verifies this is a valid probability model?
Answer: It is valid because $0.45+0.30+0.25=1.00$ . This question tests verifying a non-uniform probability model developed from observed frequencies, ensuring it meets probability rules. A non-uniform model means outcomes are not equally likely, developed by observing frequencies (100 trials: red 45, blue 30, green 25), calculating relative frequencies (0.45, 0.30, 0.25), assigning probabilities, and verifying they sum to 1 for validity. For example, P(red)=0.45, P(blue)=0.30, P(green)=0.25 sum to 1.00, confirming it's valid, while sums not equaling 1 or using raw counts would invalidate it. The correct verification is choice A, accurately stating the sum is 1.00. Common errors include miscalculating the sum like in B, using raw frequencies for the sum as in C, or misunderstanding probability bounds like D. To develop and verify: (1) collect frequencies, (2) calculate relative frequencies, (3) assign probabilities, (4) verify sum=1 and each is between 0 and 1. Non-uniform models reflect data patterns, and mistakes often involve sum errors or confusing frequencies with probabilities.
A student creates a probability model from 100 trials of a spinner: $P(\text{red})=0.45$ , $P(\text{blue})=0.30$ , $P(\text{green})=0.25$ . Which statement correctly verifies this is a valid probability model?
Answer: It is valid because $0.45+0.30+0.25=1.00$ . This question tests verifying a non-uniform probability model developed from observed frequencies, ensuring it meets probability rules. A non-uniform model means outcomes are not equally likely, developed by observing frequencies (100 trials: red 45, blue 30, green 25), calculating relative frequencies (0.45, 0.30, 0.25), assigning probabilities, and verifying they sum to 1 for validity. For example, P(red)=0.45, P(blue)=0.30, P(green)=0.25 sum to 1.00, confirming it's valid, while sums not equaling 1 or using raw counts would invalidate it. The correct verification is choice A, accurately stating the sum is 1.00. Common errors include miscalculating the sum like in B, using raw frequencies for the sum as in C, or misunderstanding probability bounds like D. To develop and verify: (1) collect frequencies, (2) calculate relative frequencies, (3) assign probabilities, (4) verify sum=1 and each is between 0 and 1. Non-uniform models reflect data patterns, and mistakes often involve sum errors or confusing frequencies with probabilities.
A science class records how often a plant's leaves are found to be dry or wet at the start of class over 50 days: Dry 32 days, Wet 18 days. Using a probability model based on this data, which is the best estimate for P(wet)?
Answer: 18/50=0.36 Out of 50 days, the leaves were wet on 18 of them, so the relative frequency for wet is 18 divided by 50, which equals 0.36, matching Choice B. Choice A, 32/50 = 0.64, is actually the probability of the leaves being dry, not wet. Choice C inverts the fraction, dividing 50 by 18 instead of 18 by 50, which does not represent a probability at all. Choice D uses the correct fraction but makes an arithmetic error, since 18 divided by 50 actually equals 0.36, not 0.18.
A weather club tracked the weather for 30 days. It rained on 12 days and was sunny on 18 days. Based on this data, what is the best probability model for the next day?
Answer: $P(\text{rain})=\frac{12}{30}=0.4,\;P(\text{sun})=\frac{18}{30}=0.6$ This question tests developing a non-uniform probability model from observed weather frequencies over 30 days, using it to predict the next day's weather. Non-uniform means outcomes like rain and sun are not equally likely, so we develop the model by observing 12 rainy days and 18 sunny days, calculating relative frequencies as 12/30=0.4 for rain and 18/30=0.6 for sun, assigning these probabilities, and verifying they sum to 1. For example, P(rain)=0.4, P(sun)=0.6, and 0.4+0.6=1.00, which is valid. The correct model uses these observed proportions. Errors might include switching the probabilities or assuming equal likelihood. Developing the model requires (1) tallying frequencies, (2) computing relative frequencies, (3) setting probabilities, and (4) checking the sum. Using the model, we can predict future events, and non-uniformity reflects actual weather patterns rather than assuming fairness.
A class uses a weighted spinner with 3 colors. In 100 practice spins, the results were: Red 45, Blue 30, Green 25. Which probability model best matches the data?
Answer: $P(\text{red})=0.45,\;P(\text{blue})=0.30,\;P(\text{green})=0.25$ This question tests developing a non-uniform probability model from observed frequencies, using it to represent the likelihood of outcomes based on data. A non-uniform model means outcomes are not equally likely, developed by observing frequencies (100 spins: red 45, blue 30, green 25), calculating relative frequencies (45/100=0.45, 30/100=0.30, 25/100=0.25), assigning these as probabilities, and verifying they sum to 1. For example, with 100 spins yielding red 45, blue 30, green 25, the probabilities are P(red)=0.45, P(blue)=0.30, P(green)=0.25, and they sum to 1.00, making it a valid model. The correct model is choice C, which accurately reflects the relative frequencies from the data. Common errors include assuming a uniform model like choice A, using raw frequencies as in B, or inventing probabilities like D without basing them on the data. To develop such a model: (1) collect outcome frequencies, (2) calculate relative frequencies by dividing by total trials, (3) assign these as probabilities, (4) verify they sum to 1. Non-uniform models arise from biased spinners or real patterns in data, and mistakes often involve assuming uniformity or not converting frequencies to probabilities between 0 and 1.
A class uses a weighted spinner with three colors. In 100 practice spins, the results were: Red 45, Blue 30, Green 25. Which probability model best matches the practice data, and does it represent a uniform or non-uniform model?
Answer: $P(R)=0.45,\;P(B)=0.30,\;P(G)=0.25$ ; non-uniform This question tests developing a non-uniform probability model from observed frequencies in spinner spins, assigning probabilities based on those frequencies, and identifying if the model is uniform or non-uniform. Non-uniform models have outcomes that are not equally likely, so to develop one, observe frequencies like 100 spins with red 45, blue 30, green 25, calculate relative frequencies as 45/100=0.45, 30/100=0.30, 25/100=0.25, assign these as probabilities, and verify they sum to 1. For example, P(red)=0.45, P(blue)=0.30, P(green)=0.25, and the sum is 1.00, which checks out, making it a valid non-uniform model since probabilities differ. The correct model is B, which uses these exact probabilities and correctly identifies it as non-uniform. Common errors include assuming a uniform model like A or D, using raw frequencies as in C, or swapping probabilities as in D. Developing a model involves (1) collecting frequencies, (2) calculating relative frequencies by dividing by total trials, (3) assigning them as probabilities, and (4) verifying the sum is 1. Non-uniform models arise from weighted or biased mechanisms, and mistakes often include assuming uniformity or not normalizing frequencies.
A student recorded the weather for 30 days: 12 rainy days and 18 sunny days. Based on these observations, what is the best probability model for the next day being rainy or sunny?
Answer: $P(\text{rain})=\frac{12}{30}=0.4$ and $P(\text{sun})=\frac{18}{30}=0.6$ This question tests developing a non-uniform probability model from observed weather frequencies over 30 days, using relative frequencies to assign probabilities for predicting the next day. Non-uniform models have outcomes not equally likely, so develop by observing 12 rainy and 18 sunny days, calculating relative frequencies as 12/30=0.4 for rain and 18/30=0.6 for sun, assigning these probabilities, and verifying they sum to 1. For example, P(rain)=0.4, P(sun)=0.6, sum=1.00, which is valid and non-uniform since probabilities differ. The correct model is A, which accurately uses these relative frequencies. Errors include assuming uniform probabilities as in B, using raw counts as in C, or swapping probabilities as in D. Developing involves (1) collecting frequencies, (2) calculating relative frequencies, (3) assigning probabilities, and (4) verifying sum=1; use the model to predict by multiplying probability by future trials. Non-uniformity comes from real patterns in data, and mistakes include not dividing by total or assuming equal likelihood.
A game uses a biased die with faces 1–6. In 120 rolls, the results were: 1: 10, 2: 14, 3: 18, 4: 22, 5: 26, 6: 30. Which statement is true?
Answer: A model from the data gives $P(6)=\frac{30}{120}=0.25$ , so the die appears non-uniform. This question tests developing a non-uniform model from biased die roll frequencies over 120 rolls and identifying non-uniformity. Non-uniform: not equal; develop by observing frequencies like 6:30, calculating P(6)=30/120=0.25, higher than 1/6≈0.167, indicating bias. For example, P(6)=0.25 vs. uniform 1/6, and increasing frequencies suggest non-uniform. The correct statement is B, recognizing the relative frequency shows non-uniformity. Errors: claiming uniform in A or C, using raw count in C, or extreme in D. Developing: (1) collect frequencies, (2) calculate relative, (3) assign, (4) verify sum=1; compare to uniform. Non-uniform from bias; mistakes: assuming fair or not calculating properly.
A teacher surveyed 100 students about their favorite after-school activity: Sports 38, Video games 34, Reading 28. A probability model is made from the survey. In a new survey of 50 students, 21 chose Sports. Which conclusion is best?
Answer: The model predicts about $0.38\times 50=19$ Sports choices, and 21 is close, so the model seems reasonable. This question tests using a non-uniform model from student activity frequencies to predict in a new survey and comparing to observations. Non-uniform: unequal; model from 100 students, P(sports)=38/100=0.38, predict 0.38×50=19, observe 21 close. For example, expected ≈19 sports, 21 is reasonably close, model fits. The correct conclusion is A, assessing fit with approximation. Errors: exact expectation in B, miscalculation in C (38 instead of 19), uniform assumption in D. Comparing: expected P×n vs. observed, assess fit; non-uniform from preferences. Mistakes: demanding exactness or math errors.
A game uses a biased die with faces 1-6. In 120 rolls, the results were: 1: 10, 2: 14, 3: 18, 4: 22, 5: 26, 6: 30. Which statement is true?
Answer: A model from the data gives P(6)=30/120=0.25, so the die appears non-uniform. Dividing the count for 6 by the total rolls gives the relative frequency: P(6) = 30/120 = 0.25, matching choice C. Since a uniform model would give each face a probability of about 1/6 (roughly 0.167), and 0.25 is clearly higher, the die appears non-uniform, not uniform as choice A claims. Choice B wrongly assumes that because 6 came up most often, it must always happen and nothing else can occur, which isn't how probability works. Choice D uses the raw count of 30 directly as if it were a probability, instead of dividing by the total number of rolls.
Students rolled two dice 80 times and recorded the sum. They found that sums of 7 appeared 18 times, while sums of 2 appeared only 1 time. If they develop probability models based on these frequencies, which statement correctly compares their observed model to the theoretical model?
Answer: Observed $P(sum = 7) = \frac{18}{80} = 0.225$ ; this is reasonably close to theoretical $P(sum = 7) = \frac{6}{36} ≈ 0.167$ The observed $P(sum = 7) = \frac{18}{80} = 0.225$ compared to theoretical $P(sum = 7) = \frac{6}{36} ≈ 0.167$ shows reasonable agreement given sampling variation. Choice A incorrectly states theoretical probability as $\frac{1}{6}$ . Choice C incorrectly claims the probabilities match when $0.0125 ≠ 0.028$ . Choice D incorrectly uses 62 as denominator and misunderstands probability calculation.
A music app tracked what genre a student listened to over 60 songs: Pop 27, Hip-hop 15, Rock 18. Using a probability model based on these frequencies, what is P(Pop)?
Answer: 27/60 = 0.45 Out of 60 songs, pop was chosen 27 times, so the relative frequency for pop is 27 divided by 60, which equals 0.45, matching Choice D. Choice A divides by 100 instead of the actual total of 60 songs, which is not the correct total to use. Choice B, 1/3, does not match the correct fraction of 27/60. Choice C flips the fraction, dividing 60 by 27 instead of 27 by 60, which does not represent a probability at all.

Start this quiz