4.112 California’s electoral college votes. During a presidential election, each state is allotted a different number of votes in the Electoral College, depending on the population. For example, California is allotted 55 votes (the most) while several states (including the District of Columbia) are allotted 3 votes each (the least). When a presidential candidate wins the popular vote in a state, the candidate wins all the Electoral College votes in that state. To become president, a candidate must win 270 of the total of 538 votes in the Electoral College. Chance(Winter 2010) demonstrated the impact on the presidential election of winning California. Assuming a candidate wins California’s 55 votes, the number of additional Electoral College votes the candidate will win can be approximated by a normal distribution with μ=241.5votes and σ=49.8votes. If a presidential candidate wins the popular vote in California, what are the chances that he or she becomes the next U.S. president?

Short Answer

Expert verified

The probability of he or she becoming the next U.S president is 0.2836.

Step by step solution

01

Given information

Assuming a candidate wins California’s 55 votes, the number of additional Electoral College votes the candidate will win can be approximated by a normal distribution with mean 241.5 votes and standard deviation 49.8 votes.

02

Calculating the probability

Let x be the random variable denoting Electoral College votes.

Here x is normally distributed with mean 241.5 votes and standard deviation 49.8 votes.

x~N(241.5,49.8)

If a presidential candidate wins the popular vote in California then the probability of he or she becoming the next U.S president has been calculated below.

To become president a candidate must have to win 270 votes.

P(x270)=1-P(x<270)=1-Px-241.549.8<270-241.549.8=1-P(z<0.572289)=1-0.716437=0.283563=0.2836P(x270)=0.2836

So, the probability of he or she becoming the next U.S president is 0.2836.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When to replace a maintenance system. An article in the Journal of Quality of Maintenance Engineering (Vol. 19,2013) studied the problem of finding the optimal replacement policy for a maintenance system. Consider a system that is tested every 12 hours. The test will determine whether there are any flaws in the system. Assume that the probability of no flaw being detected is .85. If a flaw (failure) is detected, the system is repaired. Following the fifth failed test, the system is completely replaced. Now, let x represent the number of tests until the system needs to be replaced.

a. Give the probability distribution for x as a formula.

b. Find the probability that the system needs to be replaced after 8 total tests.

4.126 Wear-out of used display panels.Wear-out failure time ofelectronic components is often assumed to have a normaldistribution. Can the normal distribution be applied to thewear-out of used manufactured products, such as coloreddisplay panels? A lot of 50 used display panels was purchasedby an outlet store. Each panel displays 12 to 18 colorcharacters. Prior to the acquisition, the panels had been usedfor about one-third of their expected lifetimes. The data inthe accompanying table (saved in the file) give the failuretimes (in years) of the 50 used panels. Use the techniquesof this section to determine whether the used panel wear-outtimes are approximately normally distributed.

0.01 1.21 1.71 2.30 2.96 0.19 1.22 1.75 2.30 2.98 0.51

1.24 1.77 2.41 3.19 0.57 1.48 1.79 2.44 3.25 0.70 1.54

1.88 2.57 3.31 0.73 1.59 1.90 2.61 1.19 0.75 1.61 1.93

2.62 3.50 0.75 1.61 2.01 2.72 3.50 1.11 1.62 2.16 2.76

3.50 1.16 1.62 2.18 2.84 3.50

Consider the discrete probability distribution shown here:

  1. Find μ=E(x).
  2. Find σ=E[(xμ)2]
  3. Find the probability that the value of x falls within one standard deviation of the mean. Compare this result to the Empirical Rule.

Voltage sags and swells. Refer to the Electrical Engineering (Vol. 95, 2013) study of the power quality of a transformer, Exercise 2.127 (p. 132). Recall that two causes of poor power quality are “sags” and “swells.”. (A sag is an unusual dip, and a swell is an unusual increase in the voltage level of a transformer.) For Turkish transformers built for heavy industry, the mean number of sags per week was 353, and the mean number of swells per week was 184. As in Exercise 2.127, assume the standard deviation of the sag distribution is 30 sags per week, and the standard deviation of the swell distribution is 25 swells per week. Also, assume that the number of sags and number of swells is both normally distributed. Suppose one of the transformers is randomly selected and found to have 400 sags and 100 swells in a week.

a. What is the probability that the number of sags per week is less than 400?

b. What is the probability that the number of swells per week is greater than 100?

4.137 The random variable xis best described by a uniform probability distribution with c= 100 and d= 200. Find the probability that xassumes a value

a. More than 2 standard deviations from

b. Less than 3 standard deviations from

c. Within 2 standard deviations of

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free