Question

A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6.3%. If the true proportion is actually 0.10, what is the probability that the sample percent will be 6.3% or less? (Round your answer(s) to 3 decimal places.)

A population consisted of 800 cases involving shoplifters. If a random sample of 250 was taken and in 85 cases the offender was under age 18, in how many of the 800 cases would you say the person involved was under age 18? (Round your answer(s) to 3 decimal places.)

Shoplifters _{(under 18)} =  cases (round your answer up to the nearest integer).

Suppose it is known that 60% of students at a particular college are smokers. A sample of 500 students from the college is selected at random. Approximate the probability that at least 270 of these students are smokers. (Round your answer(s) to 3 decimal places.) P(x  270) =

Answer 1

ANSWER:

1.

Std Dev = √[ p(1-p)/n ]   = √(0.10)(0.90)/ 75    = 0.0346

standard Z-score =   [0.063 - 0.10] / 0.0346   =   - 1.07

using standard normal Z tables, the area to the LEFT of Z = - 1.07 is   0.1423 or14.23%

2.

250/85= 800/x

x=800*85/250=272

3.

mean = np = .6*500 = 300

sd = square root of 500*.4*.6 = 10.9545

z = ( 270 - 300)/10.9656 = - 2 .74

P(Z > - 2 .74) = 1 - P(Z <- 2 .74) = 1- 0. 0031 = 0.9969