Schadek Silkscreen Printing Inc. purchases plastic cups on which to print logos for sporting events, proms, birthdays, and other special occasions. Zack Schadek, the owner, received a large shipment this morning. To ensure the quality of the shipment, he selected a random sample of 300 cups. He found 15 to be defective.
a. What is the estimated proportion defective in the population?
b. Develop a 95 percent confidenceinterval for the proportion defective. c. Zack has an agreement withhis supplier that he is to return lots that are 10 percent or moredefective. Should he return this lot? Explain your decision.

Given Information:

Number of defective cups = 15

Total number of cups = 300

Required Information:

a) defective proportion = p = ?

b) 95% confidence interval of defective proportion = ?

a) defective proportion = p = 0.05 = 5%

b) 95% confidence interval of defective proportion = (2.5%, 7.5%)

Step-by-step explanation:

The estimated defective proportion is given by

p = Number of defective cups/Total number of cups

p = 15/300

p = 0.05

p = 5%

The confidence interval of defective proportion is given by

Where p is the defective proportion, z is the z-score corresponding to 95% confidence level and n is the total number of cups.

The z-score corresponding to 95% confidence level is 1.96

CI = (2.5%, 7.5%)

So that means we are 95% confident that the defective proportion is between 2.5% to 7.5%.

Zack should not return this lot since the defective percentage is below 10%

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497