Question

Assuming that 70.1% of engineering students at Curtin University in 2007 owned an iPod or mp3 player. If you select a random sample of 100 engineering undergraduates, according to this information,

1. What is the probability that between 65% and 72% of the students own an iPod or mp3 player?
2. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage?

Please write step by step and use formula.

thank you a lot.

Answer 1

a)

Using central limit theorem,

P( < p) = P( Z < - p / sqrt( p (1 - p) / n) )

P( 0.65 < < 0.72) = P( < 0.72 ) - P( < 0.65)

= P( Z < 0.72 - 0.701 / sqrt( 0.701 * 0.299 / 100) ) < p < P( Z < 0.65 - 0.701 / sqrt( 0.701 * 0.299 / 100) )

= P( Z < 0.4150) - P( Z < -1.1140)

= 0.6609 - 0.1326

= 0.5283

b)

Symmetrical 90% limit are

Z/2 * sqrt( (1 - ) / n)

That is

0.701 1.645 * sqrt (0.701 * 0.299 / 100)

0.701 0.075

0.701 - 0.075 and 0.701 + 0.075

0.626 and 0.776

Symmetrical 90% limit are 62.6% and 76.6%