6. breakfast every day? students in an urban school were curious about how many children regularly eat…

6. breakfast every day? students in an urban school were curious about how many children regularly eat breakfast. they conducted a survey, asking, “do you eat breakfast regularly?” all 595 students in the school responded to the survey. the resulting data are summarized in the two - way table.\n| | male | female | total |\n|--|--|--|--|\n| yes | 190 | 110 | 300 |\n| no | 130 | 165 | 295 |\n| total | 320 | 275 | 595 |\nsuppose we select a student from the school at random. define event f as getting a female student and event b as getting a student who eats breakfast regularly.\na. find p(b^c). interpret this value in context.\nb. find p(female and doesnt eat breakfast regularly).\nc. find p(f or b^c).

6. breakfast every day? students in an urban school were curious about how many children regularly eat breakfast. they conducted a survey, asking, “do you eat breakfast regularly?” all 595 students in the school responded to the survey. the resulting data are summarized in the two - way table.\n| | male | female | total |\n|--|--|--|--|\n| yes | 190 | 110 | 300 |\n| no | 130 | 165 | 295 |\n| total | 320 | 275 | 595 |\nsuppose we select a student from the school at random. define event f as getting a female student and event b as getting a student who eats breakfast regularly.\na. find p(b^c). interpret this value in context.\nb. find p(female and doesnt eat breakfast regularly).\nc. find p(f or b^c).

Answer

Explanation:

Step1: Recall probability formula for complement

The formula for the probability of the complement of an event $B$ is $P(B^{C})=1 - P(B)$. The total number of students is $n = 595$, and the number of students who eat breakfast regularly $n(B)=300$. So $P(B)=\frac{n(B)}{n}=\frac{300}{595}$.

Step2: Calculate $P(B^{C})$

$P(B^{C})=1-\frac{300}{595}=\frac{595 - 300}{595}=\frac{295}{595}=\frac{59}{119}\approx0.496$. This value represents the probability that a randomly - selected student does not eat breakfast regularly.

Step3: Find $P(\text{female and doesn't eat breakfast regularly})$

The number of female students who don't eat breakfast regularly is $165$. Using the probability formula $P=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$, we have $P(\text{female and doesn't eat breakfast regularly})=\frac{165}{595}=\frac{33}{119}\approx0.277$.

Step4: Use the addition rule for probability

The addition rule for probability is $P(F\ or\ B^{C})=P(F)+P(B^{C})-P(F\ and\ B^{C})$. We know that $P(F)=\frac{275}{595}$, $P(B^{C})=\frac{295}{595}$, and $P(F\ and\ B^{C})=\frac{165}{595}$. Then $P(F\ or\ B^{C})=\frac{275 + 295-165}{595}=\frac{405}{595}=\frac{81}{119}\approx0.681$.

Answer:

a. $\frac{59}{119}\approx0.496$. The probability that a randomly - selected student does not eat breakfast regularly. b. $\frac{33}{119}\approx0.277$ c. $\frac{81}{119}\approx0.681$