AP Stats Unit 4 Practice FRQ #1

a. .211+.321+.151+.084=.767 There is a 7.67% probability that the student-athlete chosen is at least 16 years old.
b. (14*.094)+(15*.139)+… =16.548
c. This is a binomial distribution because getting a probability of 4 or 5 is either a success or failure, each student chosen is independent of each other, the number of trials is fixed at 5, and each age has a set probability of success. Binompdf(trials=5,p=.767,x=5)+Binompdf(trials=5,p=.767,x=4)=.669
The there is a 6.69% probability that 4 or 5 of the selected athletes are at least 16 years old.

Appropriate work and calculations all around. Nice job!

You’ve done appropriate & correct calculations on all three parts… but you ran yourself into trouble on parts (a) and ( c ) when you converted to a percentage. Remember to move the decimal point two spaces; you should get 76.7% and 66.8%, respectively. Unfortunately those would both bump your responses down to partial credit. Much safer is to leave your probability statements in terms of the decimal between 0 and 1 and not mention percent at all. So part (a) would say “the probability is 0.767 that the student-athlete chosen is at least 16 years old”

a) Out of the population of all the student athletes ranging from ages 14 to 19, the probability that a student selected randomly will be at least sixteen years old is solved by adding the probability that they are 16,17,18,and 19 which is respectively .211+.321+.151+.084. The probability is .767 so there is a 76.7 percent chance that a randomly selected student athlete is at least 16 years old.
b)To find the expected age of a randomly selected student athlete I would find the mean. The mean equals the summation of the ages times the probability of each age so (14 x .094)+…(19x .084) which equals 16.548, the expected age of a randomly selected student athlete. It can be a decimal since age is a continuous variable.
c) Given that the outcome is either a success(selected student athlete is at least 16) or a failure(selected student athlete is NOT at least 16), the selection of each student athlete is random, the sample size obtained is surely less than 10 percent of the total population of student athletes at this school(over 50 student athletes), and there is a set number of trial selections being 5, I will use a binomialpdf function.
n=5(trials or # of selections), p=.767(chance of success shown in part a), x=4 or 5(number of selected athletes at least 16)
For 4 student athletes : binompdf(5,.767,4) = .403
For 5 student athletes: binompdf(5,.767,5) = .265

The probability that 4/5 or 5/5 student athletes randomly selected are at least 16 is the p(4/5)+(p5/5) or .403+.265 which is .668

Good on all three parts!

(a) P(at least 16 yrs old) = P(16 yrs old) + P(17 yrs old) + P(18 yrs old) + P(19 yrs old)
= 0.211+0.321+0.151+0.084 = 0.767 or 76.7%. Therefore, the probability that the age of a student-athlete being chose from the district at random is greater than 16 is 0.767 or 76.7%.

(b) The expected age of a randomly selected student athlete is 14*(0.094) + 15*(0.139) + 16*(0.211) + 17*(0.321) + 18*(0.151) + 19*(0.084) = 16.548.

© P(4 or 5 our of 5 is at least 16 years old) = 1 - P(0,1,2,3 out of 5 is at least 16 years old)
= 1- ( addition of binomials for equal of the 4 scenarios. I will be using paper on the AP test for this)
= 0.6686

a. If a student-athlete is chosen at random, the probability that they are at least 16 is:
P(at least 16) = 0.211 + 0.321 + 0.151 +0.084 = 0.767.
b. The expected age of a randomly selected athlete is:
mu_age = 14 * 0.094 + 15 * 0.139 + 16 * 0.211 + 17 * 0.321 + 18 * 0.151 + 19 * 0.084 = 16.548 years old.
c. Let us describe the discrete random variable X as the number of athletes that are at least 16 among a group of 5 student-athletes (B(5, 0.767))
Then the probability that 4 or 5 of the selected athletes are at least 16 is:
P(X >= 4) = binomcdf (n = 5, p = 0.767, lowerbound = 4, upperbound = 5) = 0.668634

Nice work! Correct answers & calculations. I do believe that part ( c ) is unlikely to show up this year; binomial calculations are a lot by hand (but it’s clear that you’re ready just in case!)

Correct answers with appropriate work shown!

a) Let X represent the age of a student-athlete.
P(X greater than or equal to 16) = 0.211 + 0.321 + 0.151 + 0.084 = 0.767 or 76.7%

b) E(X) = 14(0.094) + 15(0.139) + 16(0.211) + 17(0.321) + 18(0.151) + 19(0.084) = 16.548 years

c) Let Y be a binomial random variable that represents the number of student athletes over 16 years of age.

P(Y=4 or Y=5) = P(Y=4) + P(Y=5) = binompdf(trials: 5; success rate: 0.767; # of successes: 4) + binompdf(trials: 5; success rate: 0.767; # of successes: 5) = 0.6686 or 66.86%

Good work! Correct answers with appropriate calculations are shown.

A) P(16yrs+17yrs+18yrs+19yrs)= .211+0.321+0.151+0084= 0.767

B) P(Expected) (14*.094) + (15*.139)+ (16* 0.211) + (17*.321) + (18 * 0.151) + (19*0.084) = 16.548 yrs

C) 0.211+0.321+0.151+0.084= 0.767
1- binomcdf( n=5, p=0.767, X=3) = 0.6686

Correct work and calculations on all three parts!

a.) The probability that a student-athlete chosen from the district at random is at least 16 years old is 0.211 + 0.321 + 0.151 + 0.084 = .767.

b.) The expected age of a randomly selected student athlete is 14(0.094) + 15(0.139) + 16(.211) +17(.321) + 18(.151) +19(.084) = 16.548.

c.)Due to the distribution having a binary output (Below age 16 and above age 16), being independent(you can only be in one individual age), having a set amount of trials (5) and having an equal probability for each student-athlete (.767), this is a binomial distribution.

This means when you do binomial cdf (5, .767, 3) and subtract that from 1 to get the opposite you get .6686

Also I had a question about the exam, do you know what percentage you’ll need to get a 4 or 5 or how many questions you need right. Thanks.

Correct answers and work for all three parts.

As for your question about exam scoring, it varies from year to year. Typically speaking if you earn around 50% of all points you’re in line for a 3, and then 4s/5s are usually around 60ish% and 70ish%. Without multiple choice it’s hard to say what the breakdown will be. But if you feel as though you can get fully correct answers on more that 2/3 of the test, you’ll likely earn a 4-5. You will not have to get all questions fully right to earn a 5.

a) P(at least 16 years old) = P(16)+P(17)+P(18)+P(19)
0.211+0.321+0.151+0.084=0.767
There is a 0.767 probability that a student-athlete chosen at random from the district is at least 16 years old.

b) 14(0.094)+15(0.139)+!6(0.211)+17(0.321)+18(0.151)+19(0.084) = 16.548
The expected age of a randomly-selected student-athlete is 16.548.

c) P(at least 16)=0.767
(5c4 * 0.767^4 * 0.233) + (0.767^5) =0.669

There is a 0.669 probability that 4 or 5 of a sample of 5 randomly selected student athletes are at least 16 years old.

Perfect! Correct answers w/ correct supporting work.

a) The probability that the randomly chosen student-athlete is at least 16 years old is 0.767.
P(x>= 16) = 0.211+0.321+0.151+0.084 = 0.767

b) The expected age or a randomly selected student athlete is 16.548 years old.
(140.094)+(150.139)+(160.211)+(170.321)+(180.151)+(190.084) = 16.548

c) The probability that 4 or 5 of the selected athletes are at least 16 years old is 0.66863.
This is a binomial probability question. n=5, p=0.767, and x=4 or 5.
binomCDF (5,0.767,4,5) = 0.66863 Followed by the sample size, probability, minimum value, and the max value.

Good - all three parts have correct calculations and supporting work

a) The probability that a student is at least 16 is .767 (.211+.321+.151+.084)
b) the expected age is 16.5 years (14(.094)+15(.139)+16(.211)+17(.321)+18(.151)+19(.084)
c) The probability that 4 out of a sample of 5 student athletes is .6686. (n=5, p=.767, summation: upper=5 lower =4, binomialpfc(5…767,X)

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