Up until 2015, the SAT had a "guessing penalty". Consider an exam with 40 questions each with six possible choices. Only one is correct. Each correct answer is worth +20 points and each incorrect one is worth -5 points. Consider answering each question by rolling a six-sided die for each question and simply choosing the corresponding answer.
What is your expected score for the first question?
[Hint: First write your score for this question as aX+b where X∼Ber(p) for some numbers a,b and p to be determined.]
What is the standard deviation of your total score on the exam? ?
What is the probability that you score more than 400 points? Use the CLT. ?
What is the probability that you score more than 600 points? Use the CLT. ?
To find the expected score for the first question, we need to determine the probability of answering it correctly and incorrectly.
Let X be a Bernoulli random variable with parameter p, representing the probability of answering the first question correctly. The value of X is 1 if the answer is correct and 0 if the answer is incorrect.
The score for the first question can be written as aX + b, where a and b are constants.
Since each correct answer is worth +20 points and each incorrect answer is worth -5 points, we have:
a = 20 (score for a correct answer)
b = -5 (score for an incorrect answer)
Now, we need to determine the value of p. Since there are six possible choices for each question and only one is correct, the probability of answering a question correctly by guessing is 1/6.
Therefore, the expected score for the first question is given by:
E(aX + b) = aE(X) + b = 20 * (1/6) + (-5) = 20/6 - 5 = 15/6 - 5.
To find the standard deviation of the total score on the exam, we need to know the number of questions on the exam. Let's assume there are n questions.
The total score on the exam can be written as S = (aX1 + b) + (aX2 + b) + ... + (aXn + b), where Xi represents the Bernoulli random variables for each question.
The mean of the total score is given by E(S) = E(aX1 + b) + E(aX2 + b) + ... + E(aXn + b) = n * (15/6 - 5).
The variance of the total score is given by Var(S) = Var(aX1 + b) + Var(aX2 + b) + ... + Var(aXn + b) = n * Var(X) = n * p * (1 - p) * a^2.
The standard deviation of the total score is then SD(S) = sqrt(Var(S)) = sqrt(n * p * (1 - p) * a^2).
To find the probability that you score more than 400 points, we can use the Central Limit Theorem (CLT) since the number of questions is large enough.
Let's assume that the number of questions is denoted by n, and the mean score per question is denoted by μ = 15/6 - 5. The standard deviation per question is given by σ = sqrt(p * (1 - p) * a^2).
The total score on the exam can be approximated as a normal distribution with mean μn = n * μ and standard deviation σn = sqrt(n) * σ.
To find the probability of scoring more than 400 points, we can calculate the z-score for 400 points using the formula:
z = (400 - μn) / σn.
Then, we can use the z-score to find the probability using a standard normal distribution table or calculator.
The same approach can be used to find the probability of scoring more than 600 points, just substitute 600 for 400 in the above calculations.