OSTP Grade 11 CCRA Statistics and Probability. Practice it free.

Students make inferences and justify conclusions from data. This Grade 11 CCRA reporting domain maps to 10 practice skills and 8 representative questions from the playable bank.

Grade 11 CCRA10 mapped skills
What the test measures

Statistics and Probability skills

  1. summarize and display data

  2. interpret models

  3. conditional probability

  4. inferential reasoning

  5. use statistics to make decisions

Standards basis

2022 Oklahoma Academic Standards for Mathematics (OAS-M) — Oklahoma

How the OSTP reports it

Oklahoma’s OSTP math is built on the 2022 Oklahoma Academic Standards for Mathematics rather than a national shared blueprint. The state math standards page notes that Statistics & Probability and Precalculus were added in the 2022 revision, and Precalculus includes a Trigonometry strand.

Blueprint & weighting

OSTP uses the Independent / state-specific framework (by grade domains structure).

Try it now

8 free questions · 0/0 correct

Practice playeasy

A survey recorded the number of text messages sent per day. Which group has the greatest standard deviation?

Group A:
50,50,50,50,5050, 50, 50, 50, 50
Group B:
40,45,50,55,6040, 45, 50, 55, 60
Group C:
48,49,50,51,5248, 49, 50, 51, 52
Group D:
30,50,50,50,7030, 50, 50, 50, 70

Group D has mean 5050 with values 3030 and 7070 each 2020 away from the center while three values sit at the mean, giving a standard deviation of about 12.612.6. Groups B and C spread less; Group A has zero spread.

Practice playeasy

The practice times, in minutes, for five days are 18,18\textsf{,} 22,22\textsf{,} 30,30\textsf{,} 20,20\textsf{,} and 35.35\textsf{.} Which data set has the same mean?

The original mean is 18+22+30+20+355=25\dfrac{18 + 22 + 30 + 20 + 35}{5} = 25. The set 15,15\textsf{,} 20,20\textsf{,} 25,25\textsf{,} 30,30\textsf{,} 3535 also has mean 2525, so the means match.

Practice playeasy

Five quiz scores are 92,92\textsf{,} 68,68\textsf{,} 75,75\textsf{,} 88,88\textsf{,} and 77.77\textsf{.} Which data set has the same median?

In order, the original values have middle value 7777, so the median is 7777. The set 70,70\textsf{,} 73,73\textsf{,} 77,77\textsf{,} 91,91\textsf{,} 9797 also has median 7777, so the medians match.

Practice playeasy

In a class, 1212 students play soccer, and 99 of those soccer players also play chess. A soccer player is picked at random. What is the probability that they also play chess?

Condition on the soccer players: P(chesssoccer)=912=34P(\text{chess}\mid\text{soccer}) = \dfrac{9}{12} = \dfrac{3}{4}.

Practice playeasy

During a fitness challenge, four teammates completed 1515, 2020, 1818, and 2222 push-ups in the first four rounds. In the fifth round, one teammate completed xx push-ups. The mean number of push-ups per round is 1919. What is the value of xx?

Five rounds with a mean of 1919 push-ups total 5×19=955 \times 19 = 95. Since 15+20+18+22=7515 + 20 + 18 + 22 = 75, x=9575=20x = 95 - 75 = 20.

Practice playeasy

A science lab uses a random molecule selector. A fixed-width release bar is divided into 44 chamber gates that should release molecules equally often. After 500500 selections, the counts were:

| Chamber | Selections |
| :-- | --: |
| A |
128128 |
| B |
8282 |
| C |
127127 |
| D |
163163 |

Which change would most likely make the selector fair?

Four equal gates should each release about 5004=125\dfrac{500}{4}=125 times (25%25\%). Chamber B was selected only 8282 times (16.4%16.4\%), so its gate is too narrow, and chamber D was selected 163163 times (32.6%32.6\%), so its gate is too wide. Widening the gate for chamber B by narrowing the gate for chamber D shifts release space from the over-used chamber to the under-used one on the shared release bar.

Practice playmedium

A grocer surveyed 2525 customers selected at random and found that 1515 of them purchased at least one organic item. Based on the survey, which of the following is the best estimate of the number of customers out of 500500 who would purchase at least one organic item?

The sample proportion is 1525=35\dfrac{15}{25} = \dfrac{3}{5}. The best estimate for 500500 customers is 35×500=300\dfrac{3}{5} \times 500 = 300 customers.

Practice playeasy

Find the mean of 5,10,155, 10, 15.

Mean =5+10+153=303=10= \dfrac{5 + 10 + 15}{3} = \dfrac{30}{3} = 10.

Keep practicing

Turn statistics and probability into game time.

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