Maine MTA High School (grade 10) Statistics and Probability. Practice it free.

Uses probability and statistical inference to analyze data and make decisions. This High School (grade 10) reporting domain maps to 12 practice skills and 8 representative questions from the playable bank.

High School (grade 10)12 mapped skills
What the test measures

Statistics and Probability skills

  1. Interpreting Categorical and Quantitative Data

  2. Making Inferences and Justifying Conclusions

  3. Conditional Probability and the Rules of Probability

  4. Using Probability to Make Decisions

Standards basis

Common Core State Standards (CCSS-M) — Maine

How the Maine MTA reports it

Maine Through-Year Assessment is Maine’s state assessment program delivered with NWEA MAP Growth item pools and through-year administration. The public sources identify it as a Maine DOE assessment aligned to Common Core State Standards in math, but Maine uses its own through-year blueprint rather than a separate national framework.

Blueprint & weighting

No official per-domain item weighting was located in the Maine DOE public assessment posts or the cited pages. The public Maine DOE materials emphasize a through-year model with fall-to-spring growth measurement rather than a published domain-by-domain blueprint.

Try it now

8 free questions · 0/0 correct

Practice playeasy

Four data sets record the number of pages read in four or five sessions. Which set has the least standard deviation?

Set A:
4,8,12,164, 8, 12, 16
Set B:
6,8,10,12,146, 8, 10, 12, 14
Set C:
7,8,9,10,117, 8, 9, 10, 11
Set D:
2,6,10,14,182, 6, 10, 14, 18

Set C has mean 99 with consecutive values from 77 to 1111, each only one or two units from the center, giving the smallest standard deviation, about 1.41.4. Sets A, B, and D spread farther from their means.

Practice playeasy

The ages, in years, of seven students are 10,10\textsf{,} 16,16\textsf{,} 14,14\textsf{,} 8,8\textsf{,} 13,13\textsf{,} 15,15\textsf{,} and 8.8\textsf{.} Which data set has the same mean?

The original mean is 10+16+14+8+13+15+87=12\dfrac{10 + 16 + 14 + 8 + 13 + 15 + 8}{7} = 12. The set 6,6\textsf{,} 8,8\textsf{,} 10,10\textsf{,} 12,12\textsf{,} 14,14\textsf{,} 16,16\textsf{,} 1818 also has mean 1212, so the means match.

Practice playeasy

Seven game scores are 90,90\textsf{,} 70,70\textsf{,} 82,82\textsf{,} 75,75\textsf{,} 95,95\textsf{,} 60,60\textsf{,} and 88.88\textsf{.} Which data set has the same median?

In order, the original values have middle value 8282, so the median is 8282. The set 63,63\textsf{,} 65,65\textsf{,} 72,72\textsf{,} 82,82\textsf{,} 88,88\textsf{,} 88,88\textsf{,} 9999 also has median 8282, so the medians match.

Practice playeasy

From a table, 5 are members who renew AND are premium and 12 total are premium members. Find the conditional probability of renewing given premium (simplified fraction).

P(AB)=P(A and B)P(B)=512=512P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{5}{12} = \dfrac{5}{12}.

Practice playeasy

A prize drawing awards $20\text{\char36}20, $10\text{\char36}10, or nothing. Winning $20\text{\char36}20 is equally likely as winning $10\text{\char36}10, and winning nothing is 33 times as likely as winning $10\text{\char36}10. What is the expected value of the prize?

Weights are 11, 11, and 33 (total 55), so probabilities are 15\dfrac{1}{5}, 15\dfrac{1}{5}, and 35\dfrac{3}{5}. Then E=piviE = \sum p_i v_i, which means the sum of each probability times its value: E=15(20)+15(10)+35(0)=$6E = \dfrac{1}{5}(20) + \dfrac{1}{5}(10) + \dfrac{3}{5}(0) = \text{\char36}6.

Practice playeasy

At trivia night, a team answered 66, 88, 77, and 99 questions correctly in the first four rounds. In the fifth round, the team answered xx questions correctly. The mean number of correct answers per round is 88. What is the value of xx?

Five rounds with a mean of 88 correct answers total 5×8=405 \times 8 = 40. Since 6+8+7+9=306 + 8 + 7 + 9 = 30, x=4030=10x = 40 - 30 = 10.

Practice playeasy

A histogram has bins 0-9: 4 values, 10-19: 7 values, 20-29: 5 values. How many data values are there in all?

Add bin frequencies: 4+7+5=164+7+5=16.

Practice playeasy

A PE class uses a spinner with 33 equal sections to assign stations: Strength, Cardio, and Flexibility. After 300300 spins, the results were:

| Station | Spins |
| :-- | --: |
| Strength |
145145 |
| Cardio |
8888 |
| Flexibility |
6767 |

Which change would most likely make the spinner fair?

Three equal sections should each be chosen about 3003=100\dfrac{300}{3}=100 times (33%33\%). Strength was chosen 145145 times (48%48\%), so its section is too large, and Flexibility was chosen only 6767 times (22%22\%), so its section is too small. Moving area from Strength to Flexibility corrects both the over- and under-represented stations.

Keep practicing

Turn statistics and probability into game time.

The Maine MTA placement starts with this test's real coverage map and finds the right difficulty.