Question

In a survey, three out of four students said that courts show too much concern for...

In a survey, three out of four students said that courts show too much concern for criminals. Find the probability that of seventy randomly selected students

At least 56 feel that courts show too much concern

Homework Answers

Answer #1

Probability that a student show too much concern for criminals

= 3/4 = 0.75

Let X be a Binomial random variable which denotes the number of students out of seventy students who show too much concern for animals

n = 70, p = 0.75

Thus , np = 52.5 and np(1-p) = 13.125

So X can be approximated to Normal distribution with

Mean = 52.5 and

standard deviation = √13.125 = 3.623

The required probability = P(X ≥ 56)

Using correction of continuity, P(X ≥ 56) = P(X > 55.5)

= P{Z > (55.5 - 52.5)/3.623}

= P(Z > 0.828)

= 0.2039

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In a survey, three out of four students said that courts show too much concern for...
In a survey, three out of four students said that courts show too much concern for criminals. Find the probability that of seventy randomly selected students At most 56 feel that courts show too much concern
In a survey, three out of four students said that courts show too much concern for...
In a survey, three out of four students said that courts show too much concern for criminals. Of seventy randomly selected students Would 68 be considered an unusually high number of students who feel that courts are partial to criminals?
in a recent poll 45% of survey respondents said that if they would prefer the child...
in a recent poll 45% of survey respondents said that if they would prefer the child to be a boy. suppose you conducted a survey of 190 randomly selected students on your campus and find that 87 of them would prefer a boy a) use the normal approximation to the binomial to approximate the probability that in a random sample of 190 students at least 87 would prefer a boy assuming the true percentage is 45% b) does this result...
In a survey of 1352 ​people, 870 people said they voted in a recent presidential election....
In a survey of 1352 ​people, 870 people said they voted in a recent presidential election. Voting records show that 62​% of eligible voters actually did vote. Given that 62​% of eligible voters actually did​ vote, (a) find the probability that among 1352 randomly selected​ voters, at least 870 actually did vote.​ (b) What do the results from part​ (a) suggest? ​(a) ​P(Xgreater than or equals870​)equals what. ​(Round to four decimal places as​ needed.)
In a survey of 1053 ​people, 780 people said they voted in a recent presidential election....
In a survey of 1053 ​people, 780 people said they voted in a recent presidential election. Voting records show that 7171​% of eligible voters actually did vote. Given that 71​% of eligible voters actually did​ vote, (a) find the probability that among 1053 randomly selected​ voters, at least 780780 actually did vote.​ (b) What do the results from part​ (a) suggest? ​(a) ​P(Xgreater than or equals≥780​)equals=nothing ​(Round to four decimal places as​ needed.)
In a survey of 11481 ​people, 742 people said they voted in a recent presidential election....
In a survey of 11481 ​people, 742 people said they voted in a recent presidential election. Voting records show that 62​% of eligible voters actually did vote. Given that 62​% of eligible voters actually did​ vote, (a) find the probability that among 1148 randomly selected​ voters, at least 742 actually did vote.​ (b) What do the results from part​ (a) suggest? ​(a)​P(X greater than or equals≥742) = ____ ​(Round to four decimal places as​ needed.)
In 2019, the National Highway Traffic Administration said that, based on survey data, 84% of drivers...
In 2019, the National Highway Traffic Administration said that, based on survey data, 84% of drivers use seat belts regularly. If four drivers are randomly selected, what is the probability that at least one of them is NOT wearing a seatbelt? (You can model this with a tree diagram.)
In a survey of 1318 ​people, 877 people said they voted in a recent presidential election....
In a survey of 1318 ​people, 877 people said they voted in a recent presidential election. Voting records show that 64​% of eligible voters actually did vote. Given that 64​% of eligible voters actually did​ vote, (a) find the probability that among 1318 randomly selected​ voters, at least 877 actually did vote.​ (b) What do the results from part​ (a) suggest?
In a survey of 1001 ​people, 689 people said they voted in a recent presidential election....
In a survey of 1001 ​people, 689 people said they voted in a recent presidential election. Voting records show that 66​% of eligible voters actually did vote. Given that 66​% of eligible voters actually did​ vote, (a) find the probability that among 1001 randomly selected​ voters, at least 689 actually did vote.​ (b) What do the results from part​ (a) suggest?
In a survey of 1368 ​people, 989 people said they voted in a recent presidential election....
In a survey of 1368 ​people, 989 people said they voted in a recent presidential election. Voting records show that 70% of eligible voters actually did vote. Given that 70​% of eligible voters actually did​ vote, (a) find the probability that among 1368 randomly selected​ voters, at least 989 actually did vote.​ (b) What do the results from part​ (a) suggest?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT