AP Stats Unit 4 Practice FRQ #1

[jerry-kosoff]

The athletics department of a large suburban school district tracks the age of student-athletes registered to play for a Varsity sport at the start of the spring season. The table below shows the distribution of age, in years, for those student-athletes.

a. If a student-athlete is chosen from the district at random, what is the probability that they are at least 16 years old?

b. Calculate the expected age of a randomly-selected student-athlete.

c. If a random sample of 5 student-athletes is selected, what is the probability that 4 or 5 of the selected athletes are at least 16 years old?

[sadierose72_57433]

(a) The probability that the student-athlete chosen from a random district is at least 16 years old is .767. (.211+.321+.151+.084)
(b) The expected age of a randomly-selected student-athlete is 16.548 years old. (14x.094)+(15x.139)+(16x.211)+(17x.321)+(18x.151)+(19x.084)
© The probability that 4 or 5 of the selected athletes are at least 16 years old is .6686. (binompdf(5,.767,4)+binompdf(5,.767,5).

[jerry-kosoff]

Hello again!

Your responses for all three parts have the correct calculations and answers (yay!). Be careful in part ( c ) with the “calculator label”: you would not earn credit for the work as you showed it if this were the real AP exam (and your answer would be scored as “partial” credit). When doing binomial calculations, you have to make it clear what n and p are, which your response does not explicitly do (you can quite literally draw arrows with n/p or put “binompdf(n = 5, p = .767, X = 4)” to remedy this). It’s also good practice to write out “binomial” somewhere in your response. I typically encourage students to put a little “side work” on their paper: something like “binomial scenario: n = 5, p = .767, X = 4 or X = 5”… and then you’ve communicated everything you need to and can put down the results from your calculator.

~Jerry

[brandon-wu]

a) P(at least 16 years old) = P(16 years old) + P(17 years old) + P(18 years old) + P(19 years old) P(at least 16 years old) = 0.211+0.321+0.151+0.084 = 0.767

There is a 0.767 probability that if a student athlete is at least 16 years old if chosen from the district at random.

b) E(x) = mux = ∑xi * Pi = (140.094)+(150.139)+(160.211)+(170.321)+(180.151)+(190.084) = 16.548 years

The expected age of a randomly-selected student-athlete is 16.548 years.

c) Binomial Distribution Conditions:
Binary - at least 16 or under 16, Independent - since you can only be in no more than one age group, the student-athletes’ ages are independent, Number of Trials = there are five students who are selected (n=5), Same Probability = probability of being at least 16 is the same for each student-athlete (p = 0.767)

X = the number of student-athletes that are at least sixteen years old where the distribution of x is binomial with p = 0.767 and n = 5

Using binomial cdf on TI Calculator with n = 5, p = 0.767, lower bound = 4, upper bound = 5, the calculated probability is 0.6686.

There is a 0.6686 probability that 4 or 5 of the selected athletes are at least 16 years old if a random sample of 5 student-athletes is selected.

[jerry-kosoff]

Hi Brandon -

All of your answers are correct, with work shown. A small thing: on part ( c ), when checking “independent”, it looks like you are making a reference to not being in more than one age group at a time… which would make the event mutually exclusive, not “independent.” Independence is when knowing the result of the first trial has no impact on future trials (which you check with your “same probability” statement). Independence can be assumed in this case because the small sample size (5) is much less than 10% of the population size (# of athletes in a “large suburban school district”), so we can assume independence even though we’re sampling w/o replacement.

[Prem]

a.) P(at least 16 years old)= 0.211+0.321+0.151+0.084=0.767
There is a 76.7% chance that if a student-athlete is chosen from the district and random then they are at least 16 years old.

b.) Expected age= (14)(0.094)+(15)(0.139)+(16)(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 years old.

c.) This scenario has each trial ends is either a success or failure, they are independent, the 4/5 selected athletes are fixed, and the probability to be at least 16 years old is the same at 0.767. Therefore, this is a binomial probability.
P(4 or 5 are at least 16)= binompdf (n=5, p=0.767, x=4)+ binompdf (n=5,p=0.767, x=5)= 0.668
There is 66.8% probability that 4 or 5 of the selected athletes are at least 16 years old if a random sample of 5 student-athletes is selected.

[jerry-kosoff]

Nicely done! Parts a, b, and c all show appropriate supporting work. One thing to be careful with on part ( c ) - in a binomial scenario, the number of trials must be fixed (in this case, n = 5), not the number of successes (you said “4 or 5 selected athletes”). This might get read as a misunderstanding of the conditions for a binomial setting, and may cost you. It’s hard for me to say for sure. With that said, your work is very good!

~Jerry

[452738_57637]

A. The probability that a randomly chosen student-athlete at least 16 years old is .767 (0.211+.321+.151+.084)

B. The expected age of a randomly-selected student is 16. To find that value I did: 14(.094)+15(.139)+16(.211)+17(.321)+18(.151)+19(.084)

C. The probability that 4 or 5 of the selected athletes are at least 16 years old is 66.8%.
To find the value I first checked the binomial experiment requirements:

1. Each trial has 2 possible outcomes
2. There’s a fix number of trials (5)
3. Outcomes are independent
4. Probability of success is the same for each trial

I then proceeded to calculate the probability by entering onto my calculator binompdf(5, .767,4) and binompdf(5,.767,5). After that I added the two values I got which ended up being .668.

[jerry-kosoff]

Welcome back -

Good work on showing your calculations on all parts. All calculations are done correctly; however it looks like in part (b) you’ve rounded your answer to the whole number 16. It’s OK for expected value to a decimal or other number that isn’t technically possible for a single individual, because expected value represents a long-run average. That is, if we randomly select many players, the average age of a randomly selected player will be about 16.548 years old - and we should leave the “.548” attached.

~Jerry

[anniexakbar]

A. The probability a randomly-selected student athlete is at least 16 years old is .211 + .321 + .151 + .084 = .767, or 76.7%.
B. The expected age of a randomly-selected student athlete is 14(.094) + 15(.139) + 16(.211) + 17(.321) + 18(.151) + 19(.084) = 16.548 years.
C. I don’t have a graphic calculator so I don’t think I can do this part haha

[yeonwoo-park]

a) The proabability that a randomly selected student athelete is at least 16 years old is 0.767. (0.211+0.321+0.151+0.084)
b) The expected age of a randomly selected student is 16.548. [14(.094)+15(.139)+16(.211)+17(.321)+18(.151)+19(.084)]
c) The situation is a binomial probability.
1 - binomcdf (n=5, p=0.767, x=3) = 0.6686

[devangana.rana]

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

b.) Expected age of a randomly-selected student-athlete: (14 * 0.094) + (150.139)+ (160.211)+ (170.321) + (180.151) + (19*0.084)= 16.548 years old.

c.) Conditions for a binomial experiment:
Binary- There are 2 possible outcomes: At least 16 or under 16.
Independent- The sample size of 5 students is less than 10% of the population (athletes in a large suburban school district).
Number of Trials- There are a set number of trials (5)
Same probability of success- Probability of being a student athlete that is at least 16 years old is 0.767 for all chosen athletes.

Calculator: P(4 or 5 of the selected athletes are at least 16 years old) = [Binompdf(n=5, p=0.767, X=5)] + [Binompdf(n=5, p=0.767, X=4] = 0.6686

QUESTION: What things will we need to write down on our paper to get full credit for this problem? Additionally, do we have to check the conditions for the binomial experiment to earn full credit? Thank you so much.

[jerry-kosoff]

Nice job on part (a) and (b)! Not to worry about part ( c ) - this kind of question likely won’t asked this year because of your very comment. (For the record, it’s a binomial situation. You may need to be able to justify why it is, but you probably won’t have to make the calculation)

~Jerry

[jerry-kosoff]

Hello again -

Good job! You’d get full credit for answers/work shown on all three parts.

~Jerry

[jerry-kosoff]

Welcome back!

Your answers to parts (a) and (b) are correct with work shown. For part ( c ), you’ve included more than you would typically need for a situation like this. On most rubrics, simply identifying the scenario as “binomial” will work (without citing all of the conditions) - as long as you clearly label the values of n, p, and X that are involved. You have done that with your calculator notation at the end. If you have time to include the rest, it doesn’t hurt, but isn’t where you need to worry

[annelise.mcdonald_54]

a) P(X >= 16 years old) --> .211+.321+.151+.084=.767 There is a 76.7% chance that a randomly selected student athlete is chosen from the district is at least 16 years old.

b) 14(.094)+15(.139)+16(.211)+17(.321)+18(.151)+19(.084)= 16.548 The expected age of a randomly selected student-athlete is 16.548 years old.

c) 5 randomly selected athletes P(X>=16) --> 1-binomcdf(5, .767, 3)
n=5 randomly selected student-athletes
p=.767 - the probability that a randomly selected student-athlete is at least 16 years old
x-1=4 out of 5 --> 4-1=3
The probability that 4 out of 5 randomly selected athletes are at least 16 years old is .668 or 66.8%.

[jerry-kosoff]

Nice job all around! Correct answers with appropriate work shown

[abar2648]

a)
p(x>=16)=0.211+0.321+0.151+0.084=0.767
Would it be p(x) or p(X) or does it not matter in this case?
b)
One-Variable Statistics tells me expected value=16.548
For this would i also need to right out the expected value equation to get E like in the (a) or is just saying I used 1-vars stats enough?
c)
binomCdf(n=5,p=0.767,lower=4,upper=5)

[jerry-kosoff]

Hmm… good question on part (a). I don’t think the difference in x or X would matter, since the variable (age) wasn’t defined with a letter. To be safe, you may want to say P(Age >= 16) as your notation. But good job showing the work. On part (b), you’ve got the right number, but you will need to show where it comes from more than the calculator. Writing out the formula is one way to do that; showing each number multiplied by each probability is another. (Many other submissions have done that, so take a look at those)

In part ( c ), you’ve shown the bare minimum for work. It’s always a good idea to say “binomial” somewhere on your paper before using the “calculator notation.” To be honest though, I don’t expect a binomial probability to show up this year (given the reliance on a calculator to solve it, and the exam is being written under the assumption that you don’t have access to one).

[ayah-abdulhaqq]

a. .211+.321+.151+.084= .767

There is a 76.7% chance that a student-athlete selected at random is at least 16 years old.

b. .094(14) + .139(15) + .211(16) + .321(17) + .151(18) + .084(19)

1.316 + 2.085 + 3.376 + 5.457 + 2.718 + 1.615=16.567

The expected age of a randomly-selected student athlete is about 16.567

c.
p( student athletes at least 16=5) = ( 5 5 ) (.767)^5 (1-.767)^5-5 = .2654

P(student athletes at least 16=4) (5 4) (.767)^4 (1-.767)^5-4 = .403

P( 4 athletes are at least 16 or 5 athletes are at least 16) =.403+.2654=.6686

There 66.86% chance that if a random sample of 5 student-athletes are selected, 4 or 5 of them are at least 16 years old.