Question

The ability to find a job after graduation is very important to GSU students as it is to the students at most colleges and universities.

Suppose we take a poll (random sample) of 3910 students classified as Juniors and find that 3281 of them believe that they will find a job immediately after graduation.

What is the 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.

(0.824, 0.854)

(0.833, 0.845)

(0.828, 0.851)

(0.829, 0.849)

use excel

A factory produces plate glass with a mean thickness of 4mm and a standard deviation of 1.1mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed.

What is the probability that the average thickness of the 100 sheets is less than 3.91 mm?  .

Round your answers to 5 decimal places.

Answer 1

Solution 1:

Given:

x = 3281

n = 3910

First we need to find sample proportion:

Confidence level = 0.99

So, level of significance = α=0.01

Zc = 2.576 ...Using excel formula, =ABS(NORMSINV(0.01/2))

99% confidence interval for population proportion is

Hence, (0.824, 0.854)

Solution 2:

Given:

Mean = = 4

Standard deviation = = 1.1

sample size = n = 100

We have to find P( < 3.91) = ...?

Using excel formula, =NORMDIST(3.91,4,1.1/SQRT(100),TRUE)

P( < 3.91) = 0.20663

Done