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\n| | male | female | total |\n|--|--|--|--|\n| yes | 190 | 110 | 300 |\n| no | 130 | 165 | 295 |\n| total | 320 | 275 | 595 |\n\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.\n\na. find p(b^c). interpret this value in context.\n\nb. find p(female and doesn’t eat breakfast regularly).\n\nc. find p(f or b^c).\n\nb. find the probability that the young adult completed high school. which probability rule did you use to find the answer?\n\nc. find the probability that the young adult has further education beyond high school. which probability rule did you use to find the answer?

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\n| | male | female | total |\n|--|--|--|--|\n| yes | 190 | 110 | 300 |\n| no | 130 | 165 | 295 |\n| total | 320 | 275 | 595 |\n\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.\n\na. find p(b^c). interpret this value in context.\n\nb. find p(female and doesn’t eat breakfast regularly).\n\nc. find p(f or b^c).\n\nb. find the probability that the young adult completed high school. which probability rule did you use to find the answer?\n\nc. find the probability that the young adult has further education beyond high school. which probability rule did you use to find the answer?

Answer

Explanation:

Step1: Recall probability - complement rule

The probability of the complement of an event $B$, $P(B^{C})=1 - P(B)$. The total number of students is $n = 595$, and the number of students who eat breakfast regularly (event $B$) is $n(B)=300$. So, $P(B)=\frac{n(B)}{n}=\frac{300}{595}$. Then $P(B^{C})=1-\frac{300}{595}=\frac{595 - 300}{595}=\frac{295}{595}=\frac{59}{119}\approx0.496$. Interpretation: The probability that a randomly - selected student does not eat breakfast regularly is approximately $0.496$ or about $49.6%$.

Step2: Find joint probability

The number of female students who do not eat breakfast regularly is $165$. The total number of students is $n = 595$. So, $P(\text{female and doesn't eat breakfast regularly})=\frac{165}{595}=\frac{33}{119}\approx0.277$.

Step3: Use addition rule for probability

The addition rule for probability is $P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)$. Here, $A = F$ (female student) and $B = B^{C}$ (student who doesn't eat breakfast regularly). $P(F)=\frac{275}{595}$, $P(B^{C})=\frac{295}{595}$, and $P(F\ and\ B^{C})=\frac{165}{595}$. Then $P(F\ or\ B^{C})=\frac{275}{595}+\frac{295}{595}-\frac{165}{595}=\frac{275 + 295-165}{595}=\frac{405}{595}=\frac{81}{119}\approx0.681$.

Answer:

a. $P(B^{C})=\frac{59}{119}\approx0.496$. The probability that a randomly - selected student does not eat breakfast regularly is about $49.6%$. b. $P(\text{female and doesn't eat breakfast regularly})=\frac{33}{119}\approx0.277$ c. $P(F\ or\ B^{C})=\frac{81}{119}\approx0.681$