Question

Suppose that public opinion in a large city is 62% in favor of increasing taxes to support the after-school programs and 38% against such an increase. If an interview of the random sample of 450 people from this city is conducted, what is the approximate probability that more than 180 of these people will be against increasing taxes?

Answer 1

P (z> (0.4 - 0.38) / SQUARE ROOT OF (0.62*0.38)/450)

Explanation:

Got it right! If you calculate this, you actually will get 19.21%, as the previous answer states.

Earned 5 points, have a lovely day!

Answer 2

19.21%

Explanation:

What we must do is calculate the z value in order to find the probability.

We have that z is equal to:

z = (x - n * p) / (n * p * q) ^ (1/2)

where x is the value to evaluate and is 180, n random sample that is 450 and p that is the probability that is 38 and q the complement of p, therefore it is 62%, we replace:

z = (180 - 450 * 0.38) / (450 * 0.38 * 0.62) ^ (1/2)

z = 0.87

increasing taxes = P (x> 180)

= P (z> 0.87)

= 1 - P (z <= 0.87)

= 1 - 0.8079

= 0.1921

In other words, the probability is 19.21%