For question 14 and 15, suppose it is well known that 78% of Americans say recyc

Question

For question 14 and 15, suppose it is well known that 78% of Americans say recycling is the best way to help the environment.

If we randomly survey 100 people, what are the mean and variance of the distribution of the sample proportion that say recycling is the best way to help the environment?

a. 0.78, 0.00172

b. 0.78, 0.00563

c. 0.88, 0.00934

d. 0.78, 0.04150

What is the probability that more than 80 people in the survey say that recycling is the best way to help the environment?

a. 0.8455

b. 0.6844

c. 0.3156

d. 0.1452

Explanation / Answer

Mean = Proportion p = 0.78

Standard deviation = sqrt (p ( 1-p) / n)

= sqrt( 0.78 * 0.22 / 100)

= 0.0415

Mean and standard deviation = 0.78 , 0.0415

Since np = 100 * 0.78 = 78 > 5 and nq = 100 * 0.22 = 22 > 5 ,

we can approximate normal distribution to binomial distribution.

P( X > x) = p( z > x - np / Sqrt( npq))

p( x > 80) = p( z > 80 - 78 / sqrt( 0.78 * 0.22 * 100)

= p( z > 0.48)

=1 - p( z < 0.48)

= 1 - 0.6844

= 0.3156

probability that more than 80 people in the survey say that recycling is the best way to help the

environment =  0.3156

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