1 / 22

STATISTICS AND PROBABILITY

STATISTICS AND PROBABILITY. CHAPTER 4. STAT. & PROBABILITY. 4.1 Sampling, Line, Bar and Circle Graphs 4.2 The Mean, Median and Mode 4.3 Counting Problems and Probability. 4.1 Unbiased Samples.

smaximo
Télécharger la présentation

STATISTICS AND PROBABILITY

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. STATISTICS AND PROBABILITY CHAPTER 4

  2. STAT. & PROBABILITY 4.1 Sampling, Line, Bar and Circle Graphs 4.2 The Mean, Median and Mode 4.3 Counting Problems and Probability

  3. 4.1 Unbiased Samples Unbiased sample is a random sample so that each member has an equal opportunity of being selected.

  4. 4.1 Example 1. A college president wants to find out which courses are popular with students. What procedure would be most appropriate for obtaining an unbiased sample of students? A. Survey a random sample of students from the English Department. B. Survey the first hundred students from an alphabetical listing. C. Survey random sample of students from list of entire student body. D. Survey random sample of students from list of entire student body.

  5. 4.1 Line and Bar Graphs 2. The graph shows the yearly average temperature from 1980 to 1985. What is the difference between the highest and lowest? 77 76 ºF 75 74 73 77 - 73 = 4 B. 77 ºF C. 1 ºF D. 4º F A. 73 ºF

  6. 4.1 Circle Graphs 6 4. The number of people employed in different work areas in a manufacturing plant are represented by the circle graph. What percent are represented in Sales and Administration combined? 3 A 5 S 3 23 Total = 40 C. 2.5% D. 7.5% A. 25% B. 20%

  7. 4.1 Relations from Data 7. Consider the following graph showing the value of a $15,000 car after 1, 2, 3, 4, 5 and 6 years. In what year did the price of the car begin to stabilize? A. 6 B. 5 C. 4 D. 3

  8. 4.1 Predictions from Data None Strong Positive Strong Negative Weak Positive Weak Negative Never select a choice that says one “caused “ the other, as the above graphs do not contain sufficient information to determine cause and effect.

  9. 4.1 Predictions from Data 9. The graph shows number of TV adds shown & number of cars sold during a 14 wk. period. Which best describes the relationship between the number of ads and cars sold? 20 Cars sold Ads 14 A. No Apparent association B. Increase in ads caused increase in sales C. Increase in sales caused increase in ads D. There is a positive association between increase in ads and increase in sales

  10. 4.2 Mean, Median & Mode Mean - sum of elements in set divided by number of elements in set. Median - middle element when arranged in order oraverage of two middle elements. Mode - most frequent element(s). If no element occurs more than once then there is no mode.

  11. 4.2 Mean, Median & Mode 1. Find mean, median & mode of the data in this sample: 6, 15, 24, 23, 29, 22, 21, 29, 29 Arrange in order: 6, 15, 21, 22, 23, 24, 29, 29, 29 A. 22, 23,29 Mode is 29 (most frequent) B.17.5, 22,29 Median is 23 (middle) C. 29, 23,22 Average too large! (6+15+21+22+23+24+29+29+29)/9 198/9=22 the mean D. 23, 22,29

  12. 4.2 Relationships & Graphs NORMAL Mean = Median = Mode SKEWED Left SKEWED Right 0 100 0 100 Mean < Med. < Mode Mode < Med.< Mean

  13. 4.2 Example 3. In a literature class, half scored 95 on a test. Most of the remaining scored 65, except for a few who scored 20. Which is true? Mean < Med. < Mode A. The mode equals the mean. Half scored 95 means mode =95 B. The median is greater than mode C. The mode is greater than mean D. The mean is greater than mode

  14. 4.2 Applications 8. The table shows the percent distribution of households by income level in 1990. What percent of the families have income of at least $35,000? 17+13+7=37 A. 47 B. 53 C. 26 D. 37

  15. 4.3 The Counting Principle To count the number of ways a sequence of events can happen, multiply the amount of ways each can occur. 1. Students are asked to rank 4 instructors from best to worst. How many different ways can the 4 instructors be ranked? 4 3 2 1 _______ x ________ x __________ x ________ 1st 2nd 3rd 4th A. 1 B. 4 C. 64 D. 24

  16. 4.3 Computing Probability It must always be the case: 0≤P(E)≤1 P(not E) = 1- P(E) P(A or B) = P(A) +P(B) - P(A and B) A and B are called mutually exclusive when P(A and B)=0 and then P(A or B) = P(A) +P(B)

  17. 4.3 Computing Probability To calculate P(A and B) P(A and B)= P(A)·P(B|A) P(B|A) is the probability of B given A has occurred. A and B are called independent events if and only if P(B|A)=P(B) and then P(A and B) = P(A)·P(B) Two events are dependent if and only if the occurrence of one event affects the probability of the other.

  18. 4.3 Example A survey at a college indicated that 90% of those taking the Essay portion of CLAST passed. If only 70% of those taking Mathpassed, what is the probability that a randomly selected student will fail both the Essay and the Math portion? Since 70% passed math, 30% or 3/10 failed and Since 90% passed essay, 10% or 1/10 failed And we will assume the two events are independent

  19. 4.3 Example A survey at a college indicated that 90% of those taking the Essay portion of CLAST passed. If only 70% of those taking Mathpassed, what is the probability that a randomly selected student will fail both the Essay and the Math portion? P(failed Math)=3/10 and P(failed Essay)=1/10 P(failed Mathand failed Essay)=(3/10)(1/10) =3/100

  20. 4.3 Example Two common sources of protein for US adults are beans & meat. If 75% of US adults eatmeat, 80% eatbeans and 70%eatbothmeat & beans, what is the probability that a randomly selected adult eats meatorbeans? P(meat or beans) =P(meat) or P(beans) - P(both) =75% + 80% - 70% = 85%

  21. Fresh. Soph. Junior Senior Male 16% 13% 10% 11% Female 14% 15% 12% 9% 4.3 Probability Application 8. The table gives the percent of students at a university by sex and student classification. Find the probability that a randomly selected student is a senior. 11% + 9% = 20%, A. 0.20 C. 0.52 D. 0.49 B. 0.30

  22. REMEMBER MATH IS FUN AND … YOU CAN DO IT

More Related