Question

Suppose that public opinion in a large city is 55% in favor of increasing taxes to support renewable energy and 45% against such an increase. If an interview of a random sample of 600 people from this city is conducted, what is the approximate probability that more than 360 of these people will be against increasing taxes?

A. 600 choose 360 times 0.55 to the 360 power times 0.45 to the 240 power.

B. Probability of z greater than quantity 0.60 minus 0.45 end quantity over square root quantity of 0.45 times 0.55 over 600 end quantity.

C. Probability of z greater than quantity 0.60 minus 0.55 end quantity over square root quantity of 0.55 times 0.45 over 600 end quantity.

D. Probability of z greater than quantity 0.60 minus 0.45 end quantity over square root quantity of 0.60 times 0.40 over 600 end quantity.

E. 600 choose 360 times 0.45 to the 360 power times 0.55 to the 240 power.

Answer 1

The approximate probability that more than 360 of these people will be against increasing taxes is P(Z> 0.6-0.45)

√0.45*0.55/600

The right answer is B.

Explanation:

According to the given data we have the following:

sample size, h=600

probability against increase tax p=0.45

The probability that in a sample of 600 people, more that 360 people will be against increasing taxes.

We find that P(P>360/600)=P(P>0.6)

The sample proposition of p is approximately normally distributed mith mean p=0.45

standard deviation σ=√P(1-P)/n=√0.45(1-0.45)/600

If x≅N(u,σ∧∧-2), then z=(x-u)/σ≅N(0,1)

Now, P(P>0.6)=P(P-P > 0.6-0.45)

σ √0.45*0.55/600

=P(Z> 0.6-0.45)

√0.45*0.55/600