CHAPTER 9 The Normal Distribution Introduction A branch of mathematics that uses probability is called statistics. Statistics is the branch of mathematics that uses observations and measurements called data to analyze, summarize, make inferences, and draw conclusions based on the data gathered. This chapter will explain some basic concepts of statistics such as measures of average and measures of variation. Finally, the relationship between probability and normal distribution will be explained in the last two sections. 147 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. Measures of Average There are three statistical measures that are commonly used for average. They are the mean, median, and mode. The mean is found by adding the data values and dividing by the number of values. EXAMPLE: Find the mean of 18, 24, 16, 15, and 12. SOLUTION: Add the values: 18 þ 24 þ 16 þ 15 þ 12 ¼ 85 Divide by the number of values, 5: 85 Ä 5 ¼ 17 Hence the mean is 17. EXAMPLE: The ages of 6 executives are 48, 56, 42, 52, 53 and 52. Find the mean. SOLUTION: Add: 48 þ 56 þ 42 þ 52 þ 53 þ 52 ¼ 303 Divide by 6: 303 Ä 6 ¼ 50.5 Hence the mean age is 50.5. The median is the middle data value if there is an odd number of data values or the number halfway between the two data values at the center, if there is an even number of data values, when the data values are arranged in order. EXAMPLE: Find the median of 18, 24, 16, 15, and 12. SOLUTION: Arrange the data in order: 12, 15, 16, 18, 24 Find the middle value: 12, 15, 16, 18, 24 The median is 16. EXAMPLE: Find the median of the number of minutes 10 people had to wait in a checkout line at a local supermarket: 3, 0, 8, 2, 5, 6, 1, 4, 1, and 0. SOLUTION: Arrange the data in order: 0, 0, 1, 1, 2, 3, 4, 5, 6, 8 The middle falls between 2 and 3; hence, the median is (2 þ 3) Ä 2 ¼ 2.5. CHAPTER 9 The Normal Distribution 148 The third measure of average is called the mode. The mode is the data value that occurs most frequently. EXAMPLE: Find the mode for 22, 27, 30, 42, 16, 30, and 18. SOLUTION: Since 30 occurs twice and more frequently than any other value, the mode is 30. EXAMPLE: Find the mode for 2, 3, 3, 3, 4, 4, 6, 6, 6, 8, 9, and 10. SOLUTION: In this example, 3 and 6 occur most often; hence, 3 and 6 are used as the mode. In this case, we say that the distribution is bimodal. EXAMPLE: Find the mode for 18, 24, 16, 15, and 12. SOLUTION: Since no value occurs more than any other value, there is no mode. A distribution can have one mode, more than one mode, or no mode. Also, the mean, median, and mode for a set of values most often differ somewhat. PRACTICE 1. Find the mean, median, and mode for the number of sick days nine employees used last year. The data are 3, 6, 8, 2, 0, 5, 7, 8, and 5. 2. Find the mean, median, and mode for the number of rooms seven hotels in a large city have. The data are 332, 256, 300, 275, 216, 314, and 192. 3. Find the mean, median, and mode for the number of tornadoes that occurred in a specific state over the last 5 years. The data are 18, 6, 3, 9, and 10. 4. Find the mean, median, and mode for the number of items 9 people purchased at the express checkout register. The data are 12, 8, 6, 1, 5, 4, 6, 2, and 6. 5. Find the mean, median, and mode for the ages of 10 children who participated in a field trip to the zoo. The ages are 7, 12, 11, 11, 5, 8, 11, 7, 8, and 6. CHAPTER 9 The Normal Distribution 149 ANSWERS 1. Mean ¼ 3 þ 6 þ 8 þ 2 þ 0 þ 5 þ Using the Normal Distribution Using the Normal Distribution By: OpenStaxCollege The shaded area in the following graph indicates the area to the left of x This area is represented by the probability P(X < x) Normal tables, computers, and calculators provide or calculate the probability P(X < x) The area to the right is then P(X > x) = – P(X < x) Remember, P(X < x) = Area to the left of the vertical line through x P(X < x) = – P(X < x) = Area to the right of the vertical line through x P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions Calculations of Probabilities Probabilities are calculated using technology There are instructions given as necessary for the TI-83+ and TI-84 calculators NOTE To calculate the probability, use the probability tables provided in [link] without the use of technology The tables include instructions for how to use them If the area to the left is 0.0228, then the area to the right is – 0.0228 = 0.9772 1/20 Using the Normal Distribution Try It If the area to the left of x is 0.012, then what is the area to the right? − 0.012 = 0.988 The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five a Find the probability that a randomly selected student scored more than 65 on the exam a Let X = a score on the final exam X ~ N(63, 5), where μ = 63 and σ = Draw a graph Then, find P(x > 65) P(x > 65) = 0.3446 The probability that any student selected at random scores more than 65 is 0.3446 Go into 2nd DISTR After pressing 2nd DISTR, press 2:normalcdf The syntax for the instructions are as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446 You get 1E99 (= 1099) by pressing 1, the EE key (a 2nd key) and then 99 Or, you can enter 10^99 instead The number 1099 is way out in the right tail of the normal curve We are calculating the area between 65 and 1099 In some instances, the lower number of the area might be –1E99 (= –1099) The number –1099 is way out in the left tail of the normal curve 2/20 Using the Normal Distribution Historical Note The TI probability program calculates a z-score and then the probability from the zscore Before technology, the z-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability In this example, a standard normal table with area to the left of the z-score was used You calculate the z-score and look up the area to the left The probability is the area to the right z= 65 – 63 = 0.4 Area to the left is 0.6554 P(x > 65) = P(z > 0.4) = – 0.6554 = 0.3446 Calculate the z-score: *Press 2nd Distr *Press 3:invNorm( *Enter the area to the left of z followed by ) *Press ENTER For this Example, the steps are 2nd Distr 3:invNorm(.6554) ENTER The answer is 0.3999 which rounds to 0.4 b Find the probability that a randomly selected student scored less than 85 b Draw a graph Then find P(x < 85), and shade the graph Using a computer or calculator, find P(x < 85) = normalcdf(0,85,63,5) = (rounds to one) The probability that one student scores less than 85 is approximately one (or 100%) c Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k) c Find the 90th percentile For each problem or part of a problem, draw a new graph Draw the x-axis Shade the area that corresponds to the 90th percentile 3/20 Using the Normal Distribution Let k = the 90th percentile The variable k is located on the x-axis P(x < k) is the area to the left of k The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher The variable k is often called a critical value k = 69.4 The 90th percentile is 69.4 This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above To get this answer on the calculator, follow this step: invNorm in 2nd DISTR invNorm(area to the left, mean, standard deviation) For this problem, invNorm(0.90,63,5) = 69.4 d Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k) d Find the 70th percentile Draw a new graph and label it appropriately k = 65.6 The 70th percentile is 65.6 This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above invNorm(0.70,63,5) = 65.6 Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three 4/20 Using the Normal Distribution Find the probability that a randomly selected golfer scored less than 65 normalcdf(1099,65,68,3) = 0.1587 A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things Suppose that the ... JOURNAL OF Veterinary Science J. Vet. Sci. (2009), 10(3), 233 238 DOI: 10.4142/jvs.2009.10.3.233 *Corresponding author Tel: +82-2-450-3670; Fax: +82-2-456-4655 E-mail: swjeong@konkuk.ac.kr The normal electroretinogram in adult healthy Shih Tzu dogs using the HMsERG June-sub Lee 1 , Kyung-hee Kim 1 , Ha-young Jang 1 , Bora Lee 1 , Joon Young Kim 2 , Soon-wuk Jeong 1, * 1 Department of Veterinary Surgery, College of Veterinary Medicine, Konkuk University, Seoul 143-701, Korea 2 Department of Veterinary Medicine, Madingley Road, Cambridge CB3 0ES, United Kingdom Electroretinography (ERG) is a reliable diagnostic tool for the diagnosis of retinal disease. It measures electric potentials occurring in the retina in response to light stimulation. In this study, we examined the normal electroretinogram using the Handheld Multispecies ERG (HMsERG) in Shih Tzu dogs. ERG recordings were performed in twelve eyes of six healthy Shih Tzu dogs. Dogs were anesthetized with a combination of medetomidine and ketamine. Proparacaine eye drops were also applied as a topical anesthetic. Tropicamide eye drops were applied for mydriasis. After 20 min of dark adaptation, we recorded the amplitudes and implicit times of the b-waves of the rod, standard rod and cone (Std R&C), high- intensity rod and cone (Hi-int R&C), and cone systems, and responses of the cones and inner retina by flicker light stimulation (cone flicker). Results showed that mean the amplitudes of a-waves of Std R&C, Hi-int R&C, and the cone responses were 141.25 μ V, 173.00 μ V, and 12.92 μ V, respectively. The b-waves of the rod responses ranged from 141.58 to 155.25 μ V; the Std R&C was 314.75 μ V, the Hi-int R&C was 329.42 μ V, the cones were 37.75 μ V, and the flicker responses were 64.08 μ V. The b/a ratios for the Std R&C, Hi-int R&C, and the cone response were 2.29, 1.94, and 3.71, respectively. Mean implicit time of the a-wave of the Std R&C was 15.12 ms, of Hi-int R&C was 13.42 ms, and of the cone response was 7.22 ms. The b-wave of the rod responses ranged from 68.12 to 72.68 ms, of Std R&C were 37.28 ms, of Hi-int R&C were 41.90, of the cone responses were 38.12 ms, and of the cone flicker responses were 22.80 ms. We believe that these parameters can be used as reference “normal” ERGs ranges for Shih Tzu dogs using the HMsERG under medetomidine and ketamine anesthesia. Keywords: electroretinogram, HMsERG, Shih Tzu dog Introduction Electroretinography (ERG) is a reliable diagnostic tool for the evaluation of retinal function. It is commonly used to diagnose retinal disorders before cataract extraction or in cases of unexplained visual loss without ophthalmoscopic abnormalities [11]. The definition of a normal range of ERG values is difficult because the results of ERG can vary due to intrinsic factors and extrinsic factors. Of the former, the eye’s state of light adaptation affects the results of the ERG recording the most, followed by species, age, transparency of the ocular media, retinal integrity, retinal circulatory disturbances, ocular opacity and pupil dilation [2]. Extrinsic factors are time and intensity for light stimulation, location and type of electrodes, kind of recording equipment used, anesthetic protocol, experimental conditions, and environment factors [2]. All of these factors may differ between various laboratories. However, yielding normal ranges for the ERG will certainly help to diagnose retinal disease processes. Normal ranges should be obtained for each specific clinic or laboratory, breed, age range, and every ERG equipment used. In this study, the amplitude and implicit time ranges of the a- and b- wave, and b/a ratio of normal eyes in healthy Shih Tzu dogs were measured using the Handheld Multispecies ERG (HMsERG) for determining the standard ERG parameters in our laboratory. Materials and Methods Animals Twelve eyes of six INTERNATIONAL JOURNAL OF ENERGY AND ENVIRONMENT Volume 6, Issue 1, 2015 pp.27-36 Journal homepage: www.IJEE.IEEFoundation.org The spatial distribution of dust sources in Iraq by using satellite images Kamal H.Lateef1, Azhaar K.Mishaal1, Ahmed M.Abud2 Ministry of Science and Technology- Renewable Energy Directorate, Iraq. Ministry of Science and Technology- Environment and Water Directorate, Iraq. Abstract Dust storms phenomenon occurs in the most regions of Iraq during the year, this paper is study this phenomenon by using the technique of satellite images, it has been used satellite images (Meteosat-9) with the sensor (SEVERI) and selected different dates of dust storms in 2012, geographic information system programs (ERDAS-GIS) has been used to discrimination the regions that cause this phenomena within the study area to prepare the images to read the real geographic coordinates and determines the regions that caused the occurrence of the dust storms represented by geographical location (Lon/Lat) and making Iraqi map describes these regions for year 2012 and compared with maps for previous years. Copyright © 2015 International Energy and Environment Foundation - All rights reserved. Keywords: Dust sources; Spatial distribution; ERDAS; Satellite images. 1. Introduction A dust storm or sand storm is a meteorological phenomenon common in arid and semi-arid regions, dust storms arise when a gust front or other strong wind blows loose sand and dirt from a dry surface. Particles are transported by saltation and suspension, a process that moves soil from one place and deposits it in another , Dust storms, one type of dust event are in most cases the result of turbulent wind [1], which raise large quantities of dust from desert surfaces and reduce visibility to less than 1km. This dust reaches concentrations in excess of 6000 µg/m3 in severe events [2]. Dust storms cause a great variety of environmental impacts. Tropsopheric aerosols, including dust, are an important component of the earth’s climate system and modify climate through their direct radiative effects of scattering and absorption [3], through indirect radiative effects via their influence on clouds microphysics [4], and by their role in processes of atmospheric chemistry [5]. According to the WMO (World Meteorological Organization) protocol, Dust events are classified according to visibility into the categories of: (1) Dust-in-Suspension: widespread dust in suspension not raised at or near the station at the time of observation; visibility is usually not greater than 10km; (2) Blowing Dust: raised dust or sand at the time of observation, reducing visibility to to 10km; (3) Dust Storm: strong winds lift large quantities of dust particles, reducing visibility to between 200 and 1000m; and (4) Severe Dust Storm: very strong winds lift large quantities of dust particles, reducing visibility to less than 200m the frequency of all dust events is [6]: ISSN 2076-2895 (Print), ISSN 2076-2909 (Online) ©2015 International Energy & Environment Foundation. All rights reserved. 28 International Journal of Energy and Environment (IJEE), Volume 6, Issue 1, 2015, pp.27-36 In the year 2011, GERIVANI submitted a paper can be help to find the impact of geological units on the wind erosion for finding dust storm sources in regions of western parts of Iran [7]. The researchers in reference [8] have calculated the dust storm velocity by determining the front pattern for the storm which are found that the velocity value is (37.62) km/h. The researchers in reference [9] found that the most important reason of the occurrence of dust storms in Iraq is the passage of a low-pressure system over Iran , the carry cool air from that region towards warmer region or warmer air of regions like eastern Syria and Iraq. 2. Materials and methods 2.1 The study area Iraq is located in south-west of Asia between (29-37 N), (39-48 E), thus it occupies the northeast corner of the Arab world, Iraq ... continuous distribution, the total area under the curve is one The parameters of the normal are the mean µ and the standard deviation σ A special normal distribution, called the standard normal distribution. .. is way out in the left tail of the normal curve 2/20 Using the Normal Distribution Historical Note The TI probability program calculates a z-score and then the probability from the zscore Before... = 0.60 The tails of the graph of the normal distribution each have an area of 0.30 Find k1, the 30th percentile and k2, the 70th percentile (0.40 + 0.30 = 0.70) 9/20 Using the Normal Distribution