In a survey, 58% of the American Adults said that they had never heard of the in

Question

In a survey, 58% of the American Adults said that they had never heard of the internet. If 20 Americans adults arc selected at random, find the probability that exactly 12 will say that they have never heard of the internet. 2d) A die is rolled 720 times. Find the mean, variance and the standard deviation of the number of 4s that will be rolled. 3a) The average daily jail population in the United States is 618, 319. If the distribution is normal and the standard deviation is 50, 200, find the probability that on a randomly selected day the jail population is: i) Greater than 700,000 ii) between 500,000 and 600,000. 3b) The national average SAT score is 1019. If we assume a normal distribution with sigma = 90, what is the 90^th percentile score? What is the probability that a randomly selected score exceeds 1200? 4a) The average annual salary in Pennsylvania was $24, 393 in 1992. Assume that salaries were normally distributed for a certain group of wage earners, and the standard deviation this group was $4362. i) Find the probability that a randomly selected individual earn less than $26,000. ii) Find the probability that, for a randomly selected sample of 25 individuals, the mean salary was less than $26.000. c) Why is the probability for b higher than the probability in part a? 4b) The average public elementary school has 458 students. Assume the standard deviation is 97. If a random sample of 36 public elementary schools is selected, find the probability that the number of students enrolled is between 450 and 465. 5a) A U.S. Travel Data Center Survey reported that Americans stayed an average of 7.5 nights when they went on a vacation. The sample size was 1500. i) Find a point estimate of the population mean. ii) Find the 95% confidence interval of the true mean. Assume the standard deviation was 0.8 night. 5b) A researcher is interested in estimating the average of teachers in the large urban school district. She wants to be 95% confident that her estimate is correct. If the standard deviation is $1050, how large a sample is needed to be accurate within $200? 5c) An irate patient complained that the cost of a doctor's visit was too high. She randomly surveyed 20 other patients and found that the mean amount of money they spent on each doctor's visit was S44.80. The standard deviation of the sample of was $3.53. i) Find a point estimate of the population mean. ii) Find the 95% confidence interval of the population mean. Assume the variable is randomly distributed.

Explanation / Answer

p = 0.58 , n= 20

f(k;n,p)=Pr(X=k)={inom {n}{k}}p^{k}(1-p)^{n-k}
P(x =12) = 20 C 12 * (0.58)^12 *(1-0.58)^(20-12)

=20C12 * 0.58^12*0.42^8

=0.1767

b) n = 720 , p = 1/6

mean = np = 720/6 = 120

variance = npq = 720*1/6*5/6 = 100

sd =sqrt(variance) = 10

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.