Curriculum for the high school grades (9-12) for the state of North Carolina (US) : Curriculum for the high school grades (9-12) for the state of North Carolina (US)
Objectives : Objectives This module explains the curriculum for the high school grades (9-12) for the state of North Carolina (US) Country: United States
State: North Carolina
Time Zone: Eastern Standard Time (GMT-5) till 11th March 2007
Subject: Mathematics
Topic: Advanced Placement Statistics
Language: English
Grade: Grade 9-12
Introduction : Students will be expected to
Observe patterns and departures from patterns.
Decide what and how to measure.
Produce models using probability and simulation.
Confirm models. Appropriate technology, from manipulatives to calculators and application software, should be used regularly for instruction and assessment. Introduction
Contents : Advanced Placement Statistics includes:
The major concepts and tools for collecting, analyzing, and drawing conclusions from data. Contents
Pre-requisites : What your student must already know?
Create and use, for sets of data, best-fit mathematical models of functions to solve problems.
Use logic and deductive reasoning to draw conclusions and solve problems.
Translate among graphic, algebraic, numeric, tabular, and verbal representations of relations.
Define and use functions to model and solve problems. Pre-requisites
Competency Goal 1 : Competency Goal 1 1.01 Summarize distributions of univariate data by determining and interpreting measures of center, spread, position, box-plots, and effects of changing units on summary measures. Sample Problem: Draw a Box-and-Whisker Plot for NATIONAL PARKS. The table gives the acreage of various national parks. Use the data to draw a box-and-whisker plot. 174,917 Parkways 594,854 Seashores 233,608 Scenic Trails 424,630 Rivers 33,431 Reserves 8,532 Memorials 228,847 Lakeshores 167,208 Historical Parks 37,557 Historic Sites 1 Battlefield Site Acreage Type of Area
Competency Goal 1 : Competency Goal 1 Step 1 Draw a number line that includes the least and greatest number in the data.
Step 2 Mark the extremes, the median, and the upper and lower quartile above the number line. Since the data have an outlier, mark the least value that is not an outlier.
Step 3 Draw the box and whiskers.
Solution:
Competency Goal 1 : 1.02 Analyze distribution of continuous univariate data (both normal and non-normal). Sample Problem: Competency Goal 1 The given frequencies can be presented in the form of cumulative percentages of the injuries 100 20 9 80 10 6 70 10 5 50 20 4 40 30 3 10 10 2 Cumulative percentages Percent of Injuries Severity of Injury 20 10 10 20 30 10 Percent of Injuries 9 6 5 4 3 2 Severity of Injury Construct a Cumulative table of the following data Solution:
Competency Goal 2 : Competency Goal 2 Sample Problem: 2.01 Construct and interpret graphical displays of univariate data In U.S. temperature is frequently reported in degrees Fahrenheit. In some other countries, temperature is reported in degrees Celsius. If the temperature in degrees Celsius can be changed to degrees Fahrenheit by using the formula y = 1.8x + 32, where y is the Fahrenheit temperature and x is the Celsius temperature. Graph y = 1.8x + 32 and use the line to find the temperature in Fahrenheit when the Celsius temperature is 30°. Solution: Using the graph of y = 1.8x + 32 the problem can be solved. A table is made with the data and a graph is drawn.
(20, 68) 68 1.8(20) + 32 20 (10, 50) 50 1.8(10) + 32 10 (0, 32) 32 1.8(0) + 32 0 (-10, 14) 14 1.8(-10) + 32 -10 (x, y) y 1.8x + 32 x
Competency Goal 2 : Competency Goal 2 Solution: (contd.) The following graph is made from the table Examining the graph, we get that when
x = 30, y = 86. Therefore, if the
temperature is 30°C, then the temperature
in Fahrenheit is 86° F.
Competency Goal 2 : Competency Goal 2 Sample Problem: 2.02 Compare distributions among sets of univariate data. Among the following data which of the ordered pairs (0, -7), (4, -10), (1, 2), or (-1, 2) are solutions of the following relation y = - 5x + 7. Make a table. Substitute the x and y values of each ordered pair into the equation.
true 12 = -5(-1) + 7
12 = 12 12 -1 true 2 = -5(1) + 7
2 = 2 2 1 false -10 = -5(4) + 7
-10 = -13 -10 4 false -7 = - 5(0) + 7
-7 = 7 -7 0 True or False? y = - 5x + 7 y x A true statement results when the ordered pairs (1, 2) and (-1, 12) are substituted into the equation. Therefore, ordered pairs (1, 2) and (-1, 12) are solutions of the equation y = - 5x + 7. Solution:
Competency Goal 3 : 3.01 Analyze categorical data. Sample Problem: Competency Goal 3 Use the table to find the mean, median, and mode of the data. 0.6 Vermont 11.4 Ohio 19.0 New York 1.3 Maine 8.2 Georgia 33.9 California 4.4 Alabama Population
(in millions) State
Competency Goal 3 : Competency Goal 3 Solution:
Competency Goal 3 : 3.02 Use and compare methods of data collection. Sample Problem: Competency Goal 3 Which measure of central tendency is most representative of the data in the table of the example of 3.01? Solution: Since there is no mode, it is to be decided whether the mean, 11.3 million, or the median, 8.2 million, is more representative of the data.
It is to be noted that the extremely large population of California greatly affected the mean.
Hence, the best representation of the data is the median, 8.2 million.
Competency Goal 3 : 3.03 Apply statistical principles and methods in sample surveys; identify difficulties. Sample Problem: Competency Goal 3 Use the table below. Find the range of the Calories. 45 Strawberries 90 Plums 84 Mandarin Oranges 113 Grapes 69 Grapefruit 84 Cherries 81 Blueberries 73 Apples Calories Fruit Calories in a 1-cup Serving of Fruit Solution: The greatest number of Calories is 113. The least number of Calories is 45. The range is 113 – 45 or
68 Calories.
Competency Goal 3 : 3.04 Apply principles and methods in designed experiments; identify difficulties. Sample Problem: Competency Goal 3 Explain which graph could be used to indicate a greater decrease in the price of gasoline?
Solution: Both graphs show that gas prices have decreased by $1.00. The bar graph looks like a $1.00 decrease, while the picture graph indicates a larger decrease. Hence the Graph B should be chosen to increase the sale.
Competency Goal 3 : 3.05 Apply concepts of probability to solve problems. Sample Problem: Competency Goal 3 Charlie Ice Cream Shoppe sells four different flavors; vanilla, chocolate, strawberry, and honey. They survey 25 customers at random. The flavors they prefer are indicated at the right. What percent of the customers prefer vanilla ice cream? 2 honey 5 strawberry 11 chocolate 7 vanilla Number Flavor Solution: 7 out of 25 customers prefer vanilla ice cream.
According to the probability, the preference of vanilla is
7 25 = 0.28
28% of the customers prefer vanilla ice cream.
Competency Goal 3 : 3.06 Use normal distributions as a model for distribution.
a) Investigate the properties of the normal distribution.
b) Use the table of standard normal distribution (Z). Sample Problem:
A home–based business found that the lengths of 5500 of their special candy canes for
the holidays are normally distributed with a mean length of 20 centimeters and a standard deviation of 0.5 centimeters.
About how many of the candy canes were between 19.5 and 20.5 centimeters in length? Competency Goal 3
Competency Goal 3 : Competency Goal 3 Solution:
Draw a normal curve. Label the mean and the mean plus or minus multiples of the standard deviation.
The values of 19.5 and 20.5 are 1
standard deviation below and above the mean, respectively. Therefore, about 68% of the data are between 19.5 and 20.5. 5500 x 68% = 3740 Multiply 5500 by 0.68.
About 3740 of the candy canes are between 19.5 and 20.5 centimeters in length.
Competency Goal 3 : 3.07 Simulate sampling distributions. Sample Problem: Competency Goal 3 You’re an operations analyst for AT&T. Long-distance telephone calls are normally distributed with = 8 min. & = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes? Solution: Followings are the limits of the normal distribution when converted from the sampling distribution The graphs depict the phenomenon more prominently
Competency Goal 3 : Competency Goal 3 Solution: (contd.) Sampling Distribution Standardized
Normal Distribution
Competency Goal 3 : 3.08 Use simulations to develop an understanding of the Central Limit Theorem and its importance in confidence intervals and tests of significance. Sample Problem: Competency Goal 3 Central Limit Theorem: Let X1, X2,..., Xn be a random sample from a population with mean and standard deviation . Let be the sample average of X1, X2,..., Xn. Then the distribution is approximately normal with mean and standard deviation . Remarks on the Central Limit Theorem, (CLT).
1. The distribution approaches a normal distribution as n gets large. In this way, the approximation improves as n increases.
2. There are probability models for which the CLT does not hold but we will not encounter these situations in this course.
3. The sum of independent (or nearly independent) random variables is key to the CLT. State Central Limit Theorem. Solution:
Competency Goal 3 : 3.09 Recognize, construct and interpret results using confidence intervals in the context of a problem. Sample Problem: Competency Goal 3 A sample of size 25 from a normal population with variance 81, produced a mean of 81.2. Find a 0.95 level of confidence interval for the mean. Solution: Here, n = 25, 2 s = 81 s Þ = 9 and x = 81.2 For 95% confidence interval for the mean we have the formula, ÷ ø ö ç è æ s + < m < s - 96 . 1 n x 96 . 1 n x P = P(81.2 – 9/5. 1.96 < m < 81.2 + 9/5. 1.96 ) = P(81.2 - 3.528 < m < 81.2 + 3.528) = P (77.672 < m < 84.728)
Competency Goal 3 : 3.10 Perform tests of significance and interpret results in the context of a problem. Sample Problem: Competency Goal 3 Solution: A sample of size 5 from a n ( 2 , s m ) population is found to have a mean - 0.1 & s.d. 1.1. Test, at 5% level of significance, the null hypothesis s = 1 against the alternative hypothisis s > 1, it be ing given that P(x 2 >9.49) = 0.05 for 4 degrees of freedom. We have to test H o : s = 1 versus H 1 : s >1 We use the statstic x = ) 1 n ( x ~ n 2 2 2 - s m For the given sample, n = 5, 1 , 1 . 1 2 2 = s = m Therefore , x = [5 X (1.1) 2 ] / 1 = 5 x 1.21 = 6.05 Also x ~ x 2 (4) under H o Since 6.05 < 9.49 i.e. the observed value is less than the critical value, the null hypothesis is accepted.
Competency Goal 4 : Competency Goal 4 4.01 Analyze bivariate data.
a) Recognize and analyze correlation and linearity.
b) Determine the least squares regression line.
c) Create residual plots and identify outliers and influential points to analyze data.
d) Use logarithmic and power transformations to analyze data. Sample Problem: Find the most likely price in New York corresponding to the price of $ 70 at California from the following data. 3.5 2.5 Standard Deviation $ 67 $ 65 Average Price New York California Cities The correlation co-efficient between the prices of commodities in the two cities is 0.8.
Competency Goal 4 : Competency Goal 4 Solution: