Question

If 10% of urban youth are arrested before they reach the age of 18, what is...

If 10% of urban youth are arrested before they reach the age of 18, what is the probability that in a sample of 100 urban children, more than 5 children will be arrested before they are 18?

0 0
Add a comment Improve this question Transcribed image text
Answer #1

x The urban the age youth are arrested of 18 before they reach of 18 IF 10 ul. urban youth are arrested before they reach ageX ~ binomial ( n ,p) X-mean = np = 1oo (01) =10 standard deviation = 6=1np q = 10o (0.10) Co.go) To=3 Here n> 50 so we calcul- - 0.0478 (from normal table) LP(x>5) = 0.9522 probability that more than 5 arrested before they are 18 children will be is

Add a comment
Know the answer?
Add Answer to:
If 10% of urban youth are arrested before they reach the age of 18, what is...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A medical researcher claims that 5% of children under 18 years of age have asthma. In...

    A medical researcher claims that 5% of children under 18 years of age have asthma. In a random sample of 300 children under 18 years of age, 24 children had asthma. Is this evidence that more than 5% of children under 18 years of age have asthma? The null and alternative Hypotheses for the Test are: HO : p = 0.05 HA : p > 0.05 The P-value for this test is 0.0086. Using a significance level of 0.10( α...

  • Thi a previous pol, 29% of adults with children under the age of 18 reported that...

    Thi a previous pol, 29% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll..300 of 1096 adults with children under the age of 18 reported that their family to dinner together seven nights a week. Is there sufficient evidence that the aroportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use...

  • statistics question about smoking before 18 years of age According to an almanac, 70% of adult smokers started smokin...

    statistics question about smoking before 18 years of ageAccording to an almanac, 70% of adult smokers started smoking before turning 18 years oh When technology is used, use the Tech Help button for further assistance. (a) Compute the mean and standard deviation of the random variable X, the number of smokers who started before 18 in 100 trials of the probability experiment (b) Interpret the mean.(c) Would it be unusual to observe 80 smokers who started smoking before turning 18 years old...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.09 hours, with a standard deviation of 2.29 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.31 hours, with a standard deviation of 1.65 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no...

  • In a previous poll, 36% of adults with children under the age of 18 reported that...

    In a previous poll, 36% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 378 of 1108 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased?...

  • 14. A random sample of 40 adults with no children under the age of 18 years results in a mean dai...

    14. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.23 hours, with a standard deviation of 2.25 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure difference in leisure time between adults with no children and adults with children (1-H2 H represent the adults with children under the age of 18 The e0% (Round...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.57 hours, with a standard deviation of 241 hours. A random sample of 40 adults with children under the age of 18 results in a meandaly leisure time of 424 hours, with a standard deviation of 1.73 hours. Construct and interpreta 90% confidence interval for the mean difference in leisure time between adults with no children and...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a meandaly leisure time of 5.92 hours, with a standard deviation of 2.33 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.26 hours, with a standard deviation of 1.61 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a meandaly leisure time of 522 hours, with a standard deviation of 2.34 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.37 hours, with a standard deviation of 1.52 hours. Construct and interpreta 95% confidence interval for the mean difference in leisure time belwoon adults with no children and...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.16 hours, with a standard deviation of 2.42 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.16 hours, with a standard deviation of 1.58 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT