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Analysis of LA Actors & Actresses: Salaries, Surgery, and Dog Preferences, Exams of Statistics

A statistics analysis of a sample of actors and actresses in los angeles. The analysis includes their salaries, the prevalence of plastic surgery and fear of aliens, and their preferred dog weights. The document also includes calculations of mean, median, standard deviation, interquartile range, and probabilities.

Typology: Exams

Pre 2010

Uploaded on 08/13/2009

koofers-user-dzc
koofers-user-dzc 🇺🇸

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ECON 230 Test 1
Name:
Section (Circle one): 9:25 10:50
You will be graded based on the answers that you write below. Please make sure to
include all of your work on the sheets provided.
Part 1:
1) B. The sample is drawn from the population of actors and actresses in Los Angeles,
so the sample can only represent that population.
2) i) Mean = 2060
ii) Median = 50
iii) Standard deviation = 4439
iv) Interquartile range = Q3 – Q1 = 5100 – 25 = 5075
v) A. Suppose that 10,000 goes up to 10,100. The new mean is 2080, but the median is
still 50 and there is still no mode.
vi) 200. Define x = sixth actor’s salary
20 30 50 200 10000 1750, so 200
6
x
Mean x
3) i) P(had plastic surgery and afraid of aliens) = 40/100 = 0.4
ii) P(afraid of aliens given that had plastic surgery) = 40/55 = 0.73
Out of the 55 aliens who have had plastic surgery, 40 are afraid of aliens.
iii) B. P(afraid of aliens) = 60/100 = 0.6
P(afraid of aliens given that had plastic surgery) = 0.73,
So P(afraid of aliens) ≠ P(afraid of aliens given that had plastic surgery)
Actors who have had plastic surgery are more likely to be afraid of aliens
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ECON 230 Test 1 Name: Section (Circle one): 9:25 10: You will be graded based on the answers that you write below. Please make sure to include all of your work on the sheets provided. Part 1:

  1. B. The sample is drawn from the population of actors and actresses in Los Angeles, so the sample can only represent that population.
  2. i) Mean = 2060 ii) Median = 50 iii) Standard deviation = 4439 iv) Interquartile range = Q 3 – Q 1 = 5100 – 25 = 5075 v) A. Suppose that 10,000 goes up to 10,100. The new mean is 2080, but the median is still 50 and there is still no mode. vi) 200. Define x = sixth actor’s salary 20 30 50 200 10000 1750, so 200 6 x Mean x
  1. i) P(had plastic surgery and afraid of aliens) = 40/100 = 0. ii) P(afraid of aliens given that had plastic surgery) = 40/55 = 0. Out of the 55 aliens who have had plastic surgery, 40 are afraid of aliens. iii) B. P(afraid of aliens) = 60/100 = 0. P(afraid of aliens given that had plastic surgery) = 0.73, So P(afraid of aliens) ≠ P(afraid of aliens given that had plastic surgery) Actors who have had plastic surgery are more likely to be afraid of aliens
  1. i)

Mean

ii) Standard deviation =

( 1.5) (80)^2 3.5 (10)^2 8.5 (10)^2

iii) A. The 50th^ percentile also happens somewhere in the first class, and 3 lbs is the only choice in the first class. iv) Lower end = 4.5 – 23.22 = -1. Upper end = 4.5 + 23.22 = 10.

  1. i) P(prefers a dog that weighs more than five pounds) = 20/100 = 0. ii) P(prefers a dog that weighs more than five pounds given that prefers a dog that weighs more than five pounds) = 10/20 = 0. Out of the 20 actors who prefer a dog that weighs over 5 pounds, 10 prefer the dog to weigh more than 10 pounds.
  2. Define Y = prefer a dog to weigh more than five pounds and N = does not prefer a dog to weigh more than five pounds You needed to be careful in each part to sample without replacement. i) P(all three actors prefer a dog to weigh more than 5 pounds) = P(YYY) = P(first actor is Y)P(second is Y|first is Y)P(third is Y|first two are Y) = 20 19 18 100 99 98

ii) Here you wanted to use the complement rule: If it’s not the case that at least one actor prefers a dog that weighs more than 5 pounds, then it is the case that no actors prefer their dog to weigh more than 5 pounds. So: P(no actors are Y) + P(at least one is Y) = 1  P(at least one is Y) = 1 - P(no actors are Y) P(no actors are Y) = P(NNN) =

So: P(at least one is Y) = 1 – 0.508 = 0. iii) We know that: P(0 actors are Y) + P(1 actor is Y) + P(2 actors are Y) + P(3 actors are Y) = 1

Part 1: You are an enterprising reporter for the Weekly World News. You hope to learn about the lives of actors and actresses. Your boss at the Weekly World News gives you a list of the phone numbers of all actors and actresses who live in Los Angeles. You pick five names at random from the list and call them up.

  1. Your sample consists of the five people you talked to. What is the population that your sample can help you to learn about? (5 points) A) All actors and actresses. B) All actors and actresses who live in Los Angeles. C) Everyone who lives in Los Angeles. D) Cannot tell from the information given.
  2. Suppose that the data below describes the salaries of the five people in your sample (in thousands of dollars): (5 points each) 20 30 50 200 10, i) What is the mean of this data? ii) Is 20 a mode for this data? iii) What is the standard deviation?

iv) What is the interquartile range? v) Suppose that the highest salary in the sample went up. A) only the mean would change B) the mean and mode would both change C) the mean and the median would change D) the mean, the median, and the mode would all change. vi) Suppose that you called up a sixth person and discovered that the mean salary of the six actors in your sample was 1750. What is the sixth actor’s salary?

Part 2

  1. After writing an article entitled “Liposuction Causes Actors to Fear Alien Invasion”, you write another article about actors’ ideal weights for their pet dogs. You ask all 100 actors: “If you had a pet dog, what is your ideal weight for that dog?” The actors’ answers are summarized in the frequency table below. (5 points each) Weight (in pounds) Frequency 1-5 80 6-10 10 11-15 10 i) Find the mean of this data. ii) True or False: Based on the data in the table, the interquartile range must be 4 pounds or less. iii) Given the frequency table, what is the most logical of the following estimates for the 50th^ percentile of this data? A) 3 lbs B) 6 lbs C) 9 lbs D) 12 lbs
  1. Suppose we pick one actor from the sample in part 4). Calculate: (5 points each) i) The probability that the actor prefers a dog that weighs more than 5 pounds. ii) The probability that the actor prefers a dog that weighs more than 15 pounds. iii) The probability that the actor prefers a dog that weighs more than 10 pounds given that the actor prefers a dog to weigh more than 5 pounds.

iv) The probability that exactly one actor prefers dogs that weigh more than 5 pounds. Extra credit: Calculate the standard deviation of the data in the table from part 4. Use the mean that you calculated in part 4) and this standard deviation, together with Chebyshev’s Theorem to find an interval that must contain 84% of the data. (3 points)