Question

researchers carried out a survey of fourth-, fifth-, and sixth-grade students in michigan. students were asked if good grades, athletic ability, or being popular was most important to them. the two - way table summarizes the survey data.\
\

\\t\\t\\t\\t\\t	4th grade	5th grade	6th grade	total	\
----	----	----	----	----	\
\multirow{4}{*}{\makecell{what is most\\\important?}}	grades	49	50	69	168	\
athletic	24	36	38	98	\
popular	19	22	28	69	\
total	92	108	135	335	\

\
suppose we select one of these students at random. whats the probability that:\
\
(a) the student is a sixth grader or a student who rated good grades as important?\
(round to 3 decimal places. leave your answer in decimal form.)\
\
(b) the student is not a sixth grader and did not rate good grades as important?\
(round to 3 decimal places. leave your answer in decimal form.)

Explanation:

Part (a)

Step1: Recall the formula for "or" probability

The formula for \( P(A \text{ or } B) \) is \( P(A) + P(B) - P(A \text{ and } B) \). Let \( A \) be the event that the student is a sixth - grader and \( B \) be the event that the student rated good grades as important.

Step2: Find \( P(A) \)

The total number of students is \( N = 335 \). The number of sixth - graders is \( n(A)=135 \). So, \( P(A)=\frac{n(A)}{N}=\frac{135}{335} \).

Step3: Find \( P(B) \)

The number of students who rated good grades as important is \( n(B) = 168 \). So, \( P(B)=\frac{n(B)}{N}=\frac{168}{335} \).

Step4: Find \( P(A \text{ and } B) \)

The number of sixth - graders who rated good grades as important is \( n(A\cap B)=69 \). So, \( P(A\cap B)=\frac{n(A\cap B)}{N}=\frac{69}{335} \).

Step5: Calculate \( P(A \text{ or } B) \)

\[

$$\begin{align*} P(A\text{ or }B)&=\frac{135}{335}+\frac{168}{335}-\frac{69}{335}\\ &=\frac{135 + 168-69}{335}\\ &=\frac{234}{335}\\ &\approx0.6985 \end{align*}$$

\]
Rounding to 3 decimal places, \( P(A\text{ or }B)\approx0.699 \)

Part (b)

Step1: Identify the event

Let \( C \) be the event that the student is not a sixth - grader and did not rate good grades as important. First, find the number of students who are not sixth - graders (\( 335 - 135=200 \)) and did not rate good grades as important. The number of students who did not rate good grades as important is \( 335 - 168 = 167 \). But we can also calculate it by looking at the table. The students who are 4th or 5th graders and chose Athletic or Popular.

The number of 4th graders who chose Athletic or Popular: \( 24 + 19=43 \)
The number of 5th graders who chose Athletic or Popular: \( 36+22 = 58 \)
So, the number of students who are not sixth - graders and did not rate good grades as important is \( 43 + 58=101 \)

Step2: Calculate the probability

The probability \( P(C)=\frac{\text{Number of students in }C}{\text{Total number of students}}=\frac{101}{335}\approx0.301 \) (rounded to 3 decimal places)

Answer:

(a) \(\boldsymbol{0.699}\)

(b) \(\boldsymbol{0.301}\)