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Chapter 17: Problem 27

Just before a referendum on a school budget, a local newspaper polls 400 voters in an attempt to predict whether the budget will pass. Suppose that the budget actually has the support of \(52 \%\) of the voters. What's the probability the newspaper's sample will lead them to predict defeat? Be sure to verify that the assumptions and conditions necessary for your analysis are met.

Short Answer

Expert verified
The probability is about 21.19%.

Step by step solution

01

Define the variables

Let \( n = 400 \) be the sample size and \( p = 0.52 \) be the actual proportion of voters supporting the budget. We need to find the probability that the poll will predict defeat, i.e., \( X/n < 0.5 \), where \( X \) is the number of voters supporting the budget in the sample.
02

Verify the conditions for normal approximation

We use the normal approximation to the binomial distribution since \( np \) and \( n(1-p) \) are large enough. Here, \( np = 400 \times 0.52 = 208 \) and \( n(1-p) = 400 \times 0.48 = 192 \), both of which are greater than 10, fulfilling the condition.
03

Calculate mean and standard deviation

The mean of the binomial distribution is \( \mu = np = 208 \), and the standard deviation is \( \sigma = \sqrt{np(1-p)} = \sqrt{400 \times 0.52 \times 0.48} \approx 9.9499 \).
04

Use the normal approximation to find the probability

We approximate \( X \) with a normal distribution \( N(208, 9.9499) \). We want the probability that \( X < 200 \) (since \( X/400 < 0.5 \) implies \( X < 200 \)). Convert to a standard normal variable: \( Z = \frac{200 - 208}{9.9499} \approx -0.804 \).
05

Find the Z-score probability

Using the standard normal distribution table, find the probability corresponding to \( Z = -0.804 \). This is approximately 0.2119.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
The binomial distribution is a probability distribution that models the number of successes in a fixed number of independent and identically distributed Bernoulli trials. Each trial has two possible outcomes: success or failure. In our exercise, each voter can either support the budget (success) or not (failure). Hence, the binomial distribution is applicable. The parameters of a binomial distribution are denoted by \( n \) and \( p \). Here, \( n = 400 \) represents the number of voters surveyed, and \( p = 0.52 \) is the probability of a voter supporting the budget. This sets up our scenario for calculating the probability of different numbers of success conditions, which in this case, is the number of voters in the sample supporting the budget.
Probability Calculation
Probability calculation involves determining the likelihood of an event occurring. For our example, we want to calculate the probability that the newspaper predicts defeat, which mathematically is when less than 50% of the voters in the sample support the budget. Using the binomial distribution, we find the probability of \( X < 200 \). However, calculating this directly with large \( n \) is cumbersome. Instead, we approximate it using the normal distribution.

The transition to the normal approximation involves checking certain conditions: both \( np \) and \( n(1-p) \) must be greater than 10. Here, \( np = 208 \) and \( n(1-p) = 192 \), both meet this criterion, allowing us to use the normal approximation. Once the normal distribution is set up with the calculated mean \( \mu = 208 \) and standard deviation \( \sigma \approx 9.9499 \), we transition to finding the probability with a Z-score.
Z-Score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is calculated in terms of standard deviations from the mean. In our case, converting the raw score to a Z-score helps us find the probability using the standard normal distribution.

The formula for the Z-score is \( Z = \frac{X - \mu}{\sigma} \). For our exercise, \( X = 200 \). So, \( Z = \frac{200 - 208}{9.9499} \approx -0.804 \). This Z-score indicates that 200 is 0.804 standard deviations below the mean of 208. Using the standard normal distribution table, we find the probability corresponding to \( Z = -0.804 \) is approximately 0.2119. This value represents the probability that fewer than 200 of the 400 voters support the budget, leading the newspaper to predict a defeat.

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Most popular questions from this chapter

A small class of five statistics students received the following scores on their AP Exam: 5,4,4,3,1 a) Calculate the mean and standard deviation of these five scores. b) List all possible sets of size 2 that could be chosen from this class. (There are \(_{5} \mathrm{C}_{2}=10\) such sets.) c) Calculate the mean of each of these sets of 2 scores and make a dotplot of the sampling distribution of the sample mean. d) Calculate the mean and standard deviation of this sampling distribution. How do they compare to those of the individual scores? Is the sample mean an unbiased estimator of the population mean?

A restaurateur anticipates serving about 180 people on a Friday evening, and believes that about \(20 \%\) of the patrons will order the chef's steak special. How many of those meals should he plan on serving in order to be pretty sure of having enough steaks on hand to meet customer demand? Justify your answer, including an explanation of what "pretty sure" means to you.

Rainfall Statistics from Cornell's Northeast Regional Climate Center indicate that Ithaca, NY, gets an average of \(35.4^{\prime \prime}\) of rain each year, with a standard deviation of \(4.2^{\prime \prime} .\) Assume that a Normal model applies. a) During what percentage of years does Ithaca get more than \(40^{\prime \prime}\) of rain? b) Less than how much rain falls in the driest \(20 \%\) of all years? c) A Cornell University student is in Ithaca for 4 years. Let \(\bar{y}\) represent the mean amount of rain for those 4 years. Describe the sampling distribution model of this sample mean, \(\bar{y}\) d) What's the probability that those 4 years average less than \(30^{\prime \prime}\) of rain?

Information on a packet of seeds claims that the germination rate is \(92 \% .\) What's the probability that more than \(95 \%\) of the 160 seeds in the packet will germinate? Be sure to discuss your assumptions and check the conditions that support your model.

According to a 2013 poll from Public Policy Polling, \(4 \%\) of American voters believe that shape-shifting reptilian people control our world by taking on human form and gaining power. Yes, you read that correctly! (This was a poll about conspiracy theories.) Assume that's the actual proportion of Americans who hold that belief. a) Use a binomial model to calculate the probability that, in a random sample of 100 people, at least \(6 \%\) of those in the sample believe the thing about reptilian people controlling our world. b) Use a Normal model to calculate the same probability. How does this compare with the answer in part a? c) That same poll found that \(51 \%\) of American voters believe there was a larger conspiracy responsible for the assassination of President Kennedy. Use a binomial model to calculate the probability that, in a random sample of 100 people, at least \(57 \%\) of those in the sample believe in the JFK conspiracy theory. d) Use a normal model to calculate the same probability. How does this compare with the answer in part c? c) What do these answers tell you about the importance of checking that \(n p\) and \(n q\) are both at least \(10 ?\)
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