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BELL RINGER. Please complete the Prerequisite skills on PG 742 #1-8. Chapter 11: Data Analysis and Statistics. Big ideas: Finding measures of central tendency and dispersion Using normal distributions Working with samples. Lesson 1: Finding Measures of Central Tendency and Dispersion.
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BELL RINGER Please complete the Prerequisite skills on PG 742 #1-8
Chapter 11:Data Analysis and Statistics Big ideas: Finding measures of central tendency and dispersion Using normal distributions Working with samples
Lesson 1: Finding Measures of Central Tendency and Dispersion
Essential Question: When will the mean of a data set differ significantly from the median of the data set?
VOCABULARY • Statistics: Numerical values used to summarize and compare sets of data • Standard deviation:The typical difference between a data value and the mean.
The data sets at the right give the waiting times (in minutes) of several people at two veterinary offices. Find the mean, median, and mode of each data set. 198 9 Office A: Mean:x 14 + 17 + + 32 … 22 = = = 9 EXAMPLE 1 Find measures of central tendency Waiting Times SOLUTION Median:20Mode:24
… 8+ 11+ + 23 144 16 = = = 9 9 Office B: Mean:x EXAMPLE 1 Find measures of central tendency Median:18Mode:18
7.4 = 1. The data set below gives the waiting times (in minutes) of 10 students waiting for a bus. Find the mean, median, and mode of the data set. 4, 8, 12, 15, 3, 2, 6, 9, 8, 7 Mean:x = … 74 4 + 8 + 12 + + 7 = 10 10 for Example 1 GUIDED PRACTICE TRANSPORTATION SOLUTION Median:7.5Mode:8
EXAMPLE 2 Find ranges of data sets Find the range of the waiting times in each data set. SOLUTION Office A: Range= 32 – 14 = 18 Office B: Range= 23 – 8 = 15 Because the range for office A is greater, its waiting times are more spread out.
The correct answer is D. ANSWER – (14 22)2 – + (17 22)2 290 Office A: = 5.7 = 9 9 ... – + + (32 22)2 ... – – (8 16)2+ (11 16)2+ + (23 16)2 – Office B: 182 = 4.5 9 = 9 EXAMPLE 3 Standardized Test Practice SOLUTION
SOLUTION Range= 15 – 2 = 13 Standard deviation = 3.8 for Examples 2 and 3 GUIDED PRACTICE • Find the range and standard deviation of the data set. • 4, 8, 12, 15, 3, 2, 6, 9, 8, 7
When will the mean of a data set differ significantly from the median of the data set? A data set with an outlier will show a greater difference between the mean and the median than will the data set with the outlier excluded
BELL RINGER: • What happens to the average when an outlier (much higher than the other values) is part of the data set?
Essential Question: Are all statistics of a data set affected when you transform the values in a data set?
VOCABULARY • Mean: The “average” of a set of numbers • Median: The middle number of a set of numbers • Mode: The number that occurs most often in a set of number • Range: A measure of dispersion equal to the difference between the greatest and least data values
NOTE: • Adding a Constant to Data Values: • The mean, median, and mode of the new data set can be obtained by adding the same constant to the mean, median and mode of the original data set. • The range and standard deviation are unchanged. • Multiplying Data Values by a Constant: • When each value of a data set is multiplied by a constant, the new mean, median, mode, range, and standard deviation can be found by multiplying each original statistic by the same constant.
Are all statistics of a data set affected when you transform the values in a data set? The range and standard deviation are not affected by an addition transformation.
EXAMPLE 1 Add a constant to data values Astronauts The data set below gives the weights (in pounds) on Earth of eight astronauts without their space suits. A space suit weighs 250 pounds on Earth.Find the mean, median, mode, range, and standard deviation of the weights of the astronauts without their space suits and with their space suits. 142, 150, 155, 156, 160, 160, 166, 175
SOLUTION EXAMPLE 1 Add a constant to data values
The data set below gives the winning distances (in meters) in the men’s Olympic triple jump events from 1964 to 2004. Find the mean, median,mode, range, and standard deviation of the distances in meters and of the distances in feet. (Note: 1 meter 3.28 feet.) EXAMPLE 2 Multiply data values by a constant OLYMPICS 16.85, 17.39, 17.35, 17.29, 17.35, 17.26, 17.61, 18.17, 18.09, 17.71, 17.79
SOLUTION EXAMPLE 2 Multiply data values by a constant
1. Astronauts: The Manned Maneuvering Unit (MMU) is equipment that latches onto an astronaut’s space suit and enables the astronaut to move around outside the spacecraft. The MMU weighs about 300 pounds on Earth. Find the mean, median, mode, range, and standard deviation of the weights of the astronauts in Example 1 with their space suits and MMUs. for Examples 1 and 2 GUIDED PRACTICE 142, 150, 155, 156, 160, 160, 166, 175
for Examples 1 and 2 GUIDED PRACTICE SOLUTION 708 Mean 708 Median 710 Mode 33 Range Standard deviation 9.3
SOLUTION 19.11 Mean 18.96 Median What If?In Example 2, find the mean, median, mode, range, and standard deviation of the distances in yards. (Note: 1 meter 1.09 yards.) 18.91 Mode 1.44 Range Standard deviation 0.40 for Examples 1 and 2 GUIDED PRACTICE 2.
Are all statistics of a data set affected when you transform the values in a data set? The range and standard deviation are not affected by an addition transformation.
BELL RINGER: • What changes when you multiply by a constant?
Essential Question: Where are the values in a normal distribution that rarely occur displayed on a normal curve?
VOCABULARY • Normal distribution: A probability distribution with mean, x and standard deviation modeled by a bell-shaped curve • Normal curve: A smooth, symmetrical, bell-shaped curve that can model normal distributions and approximate some binomial distributions • Standard normal distribution: The normal distribution with mean 0 and standard deviation 1 • Z-score:The number z of standard deviations that a data value lies above or below
A normal distribution has mean xand standard deviation σ. For a randomly selected x-value from the distribution, find P(x – 2σ ≤ x ≤ x). x x x x The probability that a randomly selected x-value lies between –2σ and is the shaded area under the normal curve shown. P(–2σ ≤ x ≤ ) EXAMPLE 1 Find a normal probability SOLUTION = 0.135 + 0.34 = 0.475
Readings higher than 200 are considered undesirable. About what percent of the readings are undesirable? b. EXAMPLE 2 Interpret normally distribute data Health The blood cholesterol readings for a group of women are normally distributed with a mean of 172 mg/dl and a standard deviation of 14 mg/dl. a. About what percent of the women have readings between 158 and 186?
a. The readings of 158 and 186 represent one standard deviation on either side of the mean, as shown below. So, 68% of the women have readings between 158 and 186. EXAMPLE 2 Interpret normally distribute data SOLUTION
b. A reading of 200 is two standard deviations to the right of the mean, as shown. So, the percent of readings that are undesirable is 2.35% + 0.15%, or 2.5%. EXAMPLE 2 Interpret normally distribute data
A normal distribution has mean and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution. x x P(≤ ) x 1. ANSWER 0.5 for Examples 1 and 2 GUIDED PRACTICE
P(> ) x 2. x ANSWER 0.5 for Examples 1 and 2 GUIDED PRACTICE
P(<< + 2σ ) x 3. x x ANSWER 0.475 for Examples 1 and 2 GUIDED PRACTICE
P( – σ<x<) x x ANSWER 0.34 for Examples 1 and 2 GUIDED PRACTICE 4.
P(x ≤ – 3σ) 5. x ANSWER 0.0015 for Examples 1 and 2 GUIDED PRACTICE
P(x > + σ) 6. x ANSWER 0.16 for Examples 1 and 2 GUIDED PRACTICE
WHAT IF?In Example 2, what percent of the women have readings between 172 and 200? 7. ANSWER 47.5% for Examples 1 and 2 GUIDED PRACTICE
EXAMPLE 3 Use a z-score and the standard normal table Biology Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each survey. The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals. Find the probability that at most 50 seals were observed during a survey.
x 50 – 73 –1.6 z = = 14.1 Use: the table to find P(x <50) P(z <– 1.6). x – EXAMPLE 3 Use a z-score and the standard normal table SOLUTION Find: the z-score corresponding to an x-value of 50. STEP 1 STEP 2 The table shows that P(z <– 1.6)= 0.0548. So, the probability that at most 50 seals were observed during a survey is about 0.0548.
EXAMPLE 3 Use a z-score and the standard normal table
ANSWER 0.8849 for Example 3 GUIDED PRACTICE 8. WHAT IF? In Example 3, find the probability that at most 90 seals were observed during a survey.
ANSWER Az-scoreof 0 indicates that thez-score and the mean are the same. Therefore, the area under the normal curve is divided into two equal parts with the mean and thez-score being equal to 0.5. for Example 3 GUIDED PRACTICE 9. REASONING: Explain why it makes sense that P(z< 0) = 0.5.
Where are the values in a normal distribution that rarely occur displayed on a normal curve? The values occur at the end of the curve
BELL RINGER: • The mean score on an exam was 78. You scored within 5 points of the mean. If x = 78 +/- 5 represents your possible score x on the exam, what is the range of your score?
Essential Question: What should be true of the sample when you conduct a survey?