chapter 2 section review day...

Chapter 2 Section Review day Notes.notebook

1

February 13, 2017

Aug 23-8:26 PM

Honors Statistics

Aug 23-8:31 PM

Daily Agenda

1. Discuss homework C2#9


2

February 13, 2017

Sep 18-12:51 PM

Oct 10-10:50 AM

Complete the SLTR worksheet

page 136-137: 1-4, 9-11

#10 use list FLIES and only do a

Normal probability plotChapter 2 Review Exercises ANSWERS IN BACK OF TEXTBOOK


3

February 13, 2017

Feb 6-12:38 PM

The distribution of St. Louis total runs is

NOT symmetric. It is skewed to the

right. It is NOT normally distributed.

Key

Feb 6-12:33 PM

x = 5.97 Sx = 3.13_

-3.42 -0.29 2.84 5.97 9.1 12.23 15.36

44

n = 58

55

58

2.84 9.14458 0.76 too

high

-0.29 12.235558 0.95 perfect

-3.42 15.36 5858

1.00 almost

perfect

First st dev off but in general pretty close


4

February 13, 2017

Feb 6-12:32 PM

IQRs = = 1.6 8-3

3.13

This value is too high, meaning that the

middle 50% of data spreads out too

many standard deviations to be

considered normally distributed.

Sep 28-8:59 AM

There is slight arcing on this normal

probability plot of the St Louis total runs,

the plot is not a straight line but it is not

terrible. The data set can be determined to

be slightly right skewed using the "line

analysis" because it falls away "right" of the

red line.


5

February 13, 2017

Feb 6-12:36 PM

This data set should not be considered Normally

Distributed.

None of the methods provide an analysis that shows the data to be normally distributed. Method 2 show the 2nd and 3rd standard deviations to be perfect but the 1st standard deviation is too large. The data is skewed but not as severely as some of the data we have analyzed.

Sep 23-11:27 AM

Book Chapter review problems: pages 136 and 137


6

February 13, 2017

Sep 23-1:28 PM

Sep 23-1:28 PM


7

February 13, 2017

Sep 23-1:28 PM

Sep 23-1:29 PM


8

February 13, 2017

Sep 23-1:29 PM

Sep 23-1:29 PM


9

February 13, 2017

Feb 21-1:33 PM

Standard deviations and percentiles 8 points

Data transformations 4 points

OGIVES 14 points

Normal curve questions 40 points

10 multiple choice questions 40 points

Thursday PART II - 20 points

using the 4 Normality methods 20 points

Feb 10-7:15 AM

Chapter 2 Review: Multiple Choice


10

February 13, 2017

Sep 22-3:20 PM

Sep 22-3:20 PM


11

February 13, 2017

Sep 22-3:22 PM

Oct 2-3:06 PM


12

February 13, 2017

Oct 2-3:07 PM

Sep 28-12:37 PM


13

February 13, 2017

Sep 29-9:20 AM

Sep 22-3:43 PM


14

February 13, 2017

Sep 22-3:43 PM

Sep 22-3:43 PM


15

February 13, 2017

Sep 22-3:43 PM

Sep 22-3:44 PM


16

February 13, 2017

Sep 22-3:44 PM

Oct 10-10:50 AM


17

February 13, 2017

Mar 3-3:45 PM

T2.1. Many professional schools require applicants to take a standardized test. Suppose that 1000 students take such a test. Several weeks after the test, Pete receives his score report: he got a 63, which placed him at the 73rd percentile. This means that

Pete did worse than about 63% of the test takers. Pete did worse than about 73% of the test takers. Pete did better than about 63% of the test takers. Pete did better than about 73% of the test takers.

Mar 3-3:49 PM

T2.2. For the Normal distribution shown, the standard deviation is closest to

5-2=3> (a) 0

> (b) 1

> (c) 2

> (d) 3

> (e) 5


18

February 13, 2017

Mar 3-3:51 PM

T2.3. Rainwater was collected in water collectors at 30 different sites near an industrial complex, and the amount of acidity (pH level) was measured. The mean and standard deviation of the values are 4.60 and 1.10, respectively. When the pH meter was recalibrated back at the laboratory, it was found to be in error. The error can be corrected by adding 0.1 pH units to all of the values and then multiplying the result by 1.2. The mean and standard deviation of the corrected pH measurements are

(4.60+0.1)(1.2) = 5.64

(1.10)(1.2) = 1.32

Mar 3-3:51 PM

T2.4. The figure shows a cumulative relative frequency graph of the number of ounces of alcohol consumed per week in a sample of 150 adults who report drinking

of these adults consume between 4 and 8 ounces per week?

60 - 20 = 40%


19

February 13, 2017

Mar 3-3:52 PM

T2.5. The average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches. If the snowfall in Chillyville exceeds 60 inches in 15% of the years, what is the standard deviation?

4.83 inches 5.18 inches 6.04 inches 8.93 inches The standard deviation cannot be computed from the given

information.

Sep 29-12:12 PM


20

February 13, 2017

Mar 3-3:53 PM

T2.6. The figure shown is the density curve of a distribution. Seven values are marked on the density curve. Which of the following statements is true?

The mean of the distribution is E. The area between B and F is 0.50. The median of the distribution is C. The 3rd quartile of the distribution is

D.

Mar 3-3:53 PM

T2.7. If the heights of a population of men follow a Normal distribution, and 99.7% have heights between 5′0″ and 7′0″, what is your estimate of the standard deviation of the heights in this population?

1 foot = 12 inches

12/3 = 4 inches


21

February 13, 2017

Mar 3-3:54 PM

T2.8. Which of the following is correct about a standard Normal distribution?

The proportion of scores that satisfy 0 < < 1.5 is 0.4332. The proportion of scores that satisfy < −1.0 is 0.1587. The proportion of scores that satisfy > 2.0 is 0.0228. The proportion of scores that satisfy < 1.5 is 0.9332. The proportion of scores that satisfy > −3.0 is 0.9938.

Mar 3-3:54 PM

T2.9. What is the standardized score ( -score) for a

Questions T2.9 and T2.10 refer to the following setting. Until the scale was changed in 1995, SAT scores were based on a scale set many years ago. For Math scores, the mean under the old scale in the 1990s was 470 and the standard deviation was 110. In 2009,the mean was 515 and the standard deviation was 116.

z = _500-470 = 0.27110


22

February 13, 2017

Mar 3-3:55 PM

Questions T2.9 and T2.10 refer to the following setting. Until the scale was changed in 1995, SAT scores were based on a scale set many years ago. For Math scores, the mean under the old scale in the 1990s was 470 and the standard deviation was 110. In 2009,the mean was 515 and the standard deviation was 116.

T2.10. Gina took the SAT in 1994 and scored 500. Her cousin Colleen took the

Colleen–she scored 30 points higher than Gina. Colleen–her standardized score is higher than Gina’s. Gina–her standardized score is higher than Colleen’s. Gina–the standard deviation was bigger in 2013. The two cousins did equally well–their -scores are the same.

z = _500-470 = 0.27110

z = _530-515 = 0.129116

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