a.

To find: The percent of the vote does the sample suggest this candidate will get.

According to the sample, candidate will get 52% of votes.

Given information:

Confidence level: 95%

In a random sample of 625 registered voters, 325 of them express a preference for a candidate.

Calculation:

To compute the percentage of the vote the candidate will get by using sample, divide 325 by 625 multiply by 100.

325625×100=52%

The required percentage is 52%.

b.

To find: The standard deviation of the number of supporters that pollsters might find in

samples this size.

The standard deviation is 12.49.

Given information:

Confidence level: 95%

In a random sample of 625 registered voters, 325 of them express a preference for a candidate.

Calculation:

Let p be the proportion of votes to the candidate.

The estimated value for p is 0.52, then q=0.48 .

The computation of standard deviation is shown.

σ=npq=625×0.52×0.48≈12.49

Hence, the standard deviation is 12.49.

c.

To verify : A normal model is appropriate here.

It is verified that a normal model is appropriate here.

Given information:

Confidence level: 95%

In a random sample of 625 registered voters, 325 of them express a preference for a candidate.

Concept used:

If we expect at least 10 successes and 10 failures (np≥10 and nq≥10) , then the binomial distribution for the number of successes can be approximated by a Normal model with Mean =np and standard deviation=npq .

Explanation:

Since

np=625×0.52=325

And

nq=300 , it means both successes and failures are greater than 10.

Thus, it can follow normal distribution.

d.

To find : The margin of error.

The margin of error is about 4%.

Given information:

Confidence level: 95%

In a random sample of 625 registered voters, 325 of them express a preference for a candidate.

Explanation:

The z -value for 95% confidence interval is 1.96.

The computation of margin of error is shown.

E=z×pqn=1.96⋅0.52⋅0.48625≈±4%

e.

To explain : The pollsters might say this election is “too close to call.”

Since the margin of error is so small implies results from sample are closer to the reality.

Given information:

Confidence level: 95%

In a random sample of 625 registered voters, 325 of them express a preference for a candidate.

Explanation:

Lesser the margin of error, more the confidence on the results.

Since the margin of error is so small implies results from sample are closer to the reality.

Chapter 10.4, Problem 54E

Chapter 10.4, Problem 56E

Chapter 10 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

Ch. 10.1 - Prob. 1QR Ch. 10.1 - Prob. 2QR Ch. 10.1 - Prob. 3QR Ch. 10.1 - Prob. 4QR Ch. 10.1 - Prob. 5QR Ch. 10.1 - Prob. 6QR Ch. 10.1 - Prob. 7QR Ch. 10.1 - Prob. 8QR Ch. 10.1 - Prob. 9QR Ch. 10.1 - Prob. 10QR

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.1 - Prob. 33E Ch. 10.1 - Prob. 34E Ch. 10.1 - Prob. 35E Ch. 10.1 - Prob. 36E Ch. 10.1 - Prob. 37E Ch. 10.1 - Prob. 38E Ch. 10.1 - Prob. 39E Ch. 10.1 - Prob. 40E Ch. 10.1 - Prob. 41E Ch. 10.1 - Prob. 42E Ch. 10.1 - Prob. 43E Ch. 10.1 - Prob. 44E Ch. 10.1 - Prob. 45E Ch. 10.1 - Prob. 46E Ch. 10.1 - Prob. 47E Ch. 10.1 - Prob. 48E Ch. 10.1 - 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