Question
In the table above, what are \(\mathrm{P}(\mathrm{A}\) and \(\mathrm{E})\) and \(\mathrm{P}(\mathrm{C} \mid \mathrm{E})\) ?
(a) \(12 / 125,28 / 125\)
(b) \(12 / 63,28 / 60\)
(c) \(12 / 125,28 / 63\)
(d) \(12 / 125,28 / 60\)
(e) \(12 / 63,28 / 63\)
▶️Answer/Explanation
Ans:The correct answer is (c).
There are 12 values in the \(A\) and \(E\) cell out of the total of 125 . When we are given column \(\mathrm{E}\), the total is 63 . Of those, 28 are \(\mathrm{C}\).
Question
For the tree diagram pictured above, what is P(B|X)?
(a) 1/4
(b) 5/17
(c) 2/5
(d) 1/3
(e) 4/5
▶️Answer/Explanation
Ans:
The correct answer is (b).
$
\begin{gathered}
P(\mathrm{X})=(0.8)(0.3)+(0.2)(0.5)=0.34 \\
P(\mathrm{~B} \mid \mathrm{X})=\frac{(0.2)(0.5)}{(0.8)(0.3)+(0.2)(0.5)}=\frac{0.10}{0.34}=\frac{5}{17} .
\end{gathered}
$
(This problem is an example of what is known as Bayes’s rule. It’s still conditional probability, but sort of backwards. That is, rather than being given a path and finding the probability of going along that path \(-P(X \mid B)\) refers to the probability of first traveling along \(B\) and then along \(X\)-we are given the outcome and asked for the probability of having gone along a certain path to get there- \(P(B \mid X)\) refers to the probability of having gotten to \(X\) by first having traveled along B. You don’t need to know Bayes’s rule by name for the AP exam, but you may have to solve a problem like this one.)
Question
It turns out that 25 seniors at Fashionable High School took both the AP Statistics exam and the AP Spanish Language exam. The mean score on the Statistics exam for the 25 seniors was 2.4 with a standard deviation of 0.6 , and the mean score on the Spanish Language exam was 2.65 with a standard deviation of 0.55 . We want to combine the scores into a single score. What are the correct mean and standard deviation of the combined scores?
(a) \(5.05 ; 1.15\)
(b) \(5.05 ; 1.07\)
(c) \(5.05 ; 0.66\)
(d) \(5.05 ; 0.81\)
(e) 5.05; you cannot determine the standard deviation from this information.
▶️Answer/Explanation
Ans:The correct answer is (e).
If you knew that the variables “Score on Statistics Exam” and “Score on Spanish Language Exam” were independent, then the standard deviation would be given by
$
\sqrt{\sigma_1^2+\sigma_2^2}=\sqrt{(0.6)^2+(0.55)^2} \approx 0.82 \text {. }
$
However, you cannot assume that they are independent in this situation. In fact, they aren’t because we have two scores on the same people. Hence, there is not enough information.
Question
The GPAs (grade point averages) of students who take the AP Statistics exam are approximately normally distributed with a mean of 3.4 and a standard deviation of 0.3 . Using Table A, what is the probability that a student selected at random from this group has a GPA lower than 3.0?
(a) 0.0918
(b) 0.4082
(c) 0.9082
(d) -0.0918
(e) 0
▶️Answer/Explanation
Ans: The correct answer is (a).
$
P(X<3.0)=P\left(z<\frac{3-3.4}{0.3}=-1.33\right)=0.0918 .
$
The calculator answer is normalcdf(-100,3,3.4,0.3) \(=0.0912\). Note that answer (d) makes no sense since probability values must be nonnegative (and, of course, less than or equal to 1 ).
Question
The 2000 Census identified the ethnic breakdown of the state of California to be approximately as follows: White: \(46 \%\), Latino: \(32 \%\), Asian: \(11 \%\), Black: \(7 \%\), and Other: \(4 \%\). Assuming that these are mutually exclusive categories (this is not a realistic assumption), what is the probability that a randomly selected person from the state of California is of Asian or Latino descent?
(a) \(46 \%\)
(b) \(32 \%\)
(c) \(11 \%\)
(d) \(43 \%\)
(e) \(3.5 \%\)
▶️Answer/Explanation
Ans:
The correct answer is (d).
Because ethnic group categories are assumed to be mutually exclusive, \(\mathrm{P}(\) Asian or Latino \()=\mathrm{P}(\) Asian \()+\mathrm{P}(\) Latino \()=32 \%+11 \%=\) \(43 \%\).