ST 201D: Activity 3 The Normal Distribution Due Monday February 13th at 11:59PM PST 40 Points Overview The goal of this activity is to explore scenarios concerning the normal distribution. There are two parts of this activity, the first part is designed for you watch a video example and to further practice problems using the normal distribution. The second part is a graded assessment. Concepts are covered in the notes and examples from week five and chapter 3 in the textbook. Please review these materials prior to attempting this activity. Learning Outcomes After this activity you should be able to: Draw a normal distribution based on a given mean and standard deviation. Label the mean and standard deviation. Define the standard normal distribution and explain how it is related to general normal distributions. Approximate proportions/vales based on the 68-95-99.7 rule. Find proportions/probabilities of normally distributed random variables, using the standard normal table or TI-84. Find a value associated with that proportion (percentile), using the standard normal table or TI-84. Materials Needed TI-84 Calculator or Standard Normal Table \"Z table\" Scanner to upload Activity Part I: Watch example videos in week 5 module. Follow the instructions to complete the Normal distribution \"MCAT\" practice question. Part II: Graded Portion. Upload your completed activity by the due date into the Activity 3 link on Canvas contained in the Week 5 module. Please only upload the graded portion of the activity pages as a .pdf or word doc. You may neatly hand write or type your answers. Feel free to discuss activity with class members however your final solutions should be your own! Duplicate activities will be considered as cheating and students involved will be reported in violation of student conduct. 1 Part I: Normal Distributions The total scores on the Medical College Admission Test (MCAT) are known to follow a Normal distribution. In 2010, the mean MCAT score for all who took the exam was 25.0 with a standard deviation of 6.4. 1) What is the population of interest? All persons who took the MCAT exam in 2010 2) What are the population parameters for this problem? = \"mu\" is the population mean = 25.0 = \"sigma\" is the population standard deviation = 6.4 3) What does the distribution look like? Let's draw it! The standard normal distribution will be bell-shaped and symmetric with the mean as the center value. From here you can calculate and draw the values associated with 1 , 2 and 3 . That is, the values that represent one standard deviation away from the mean, two standard deviations away from the mean, and three standard deviations away from the mean. 4) What percentage of people who took the MCAT in 2010 scored below 25? Given this is an exact normal distribution the mean and the median will always be the same, because 50% of the data falls below the median we then know that 50% of the data falls below the mean which is 25. The percentage (or proportion) of people who took the MCAT in 2010 who scored below 25 is 50% (or .50). 5) What percentage of people who took the MCAT in 2010 scored between 12.2 and 95% Why? The 68-95-99.7 rule tells us that the proportion or percentage of values between 2 standard deviations from the mean will be 95%. 37.8? 2 6) What percentage of people who took the MCAT in 2010 scored above 37.8? 2.5% Why? The 68-95-99.7 rule tells us that the proportion or percentage of values between 2 standard deviations from the mean will be 95%. The score of 37.8 represents +2 , if we want to know who scored more than this than we will need to know the area that is greater than 37.8. This is simply Areathe outer tails 100 95 5 = = = only the upper tail = 2.5% 2 2 2 7) What proportion of students taking the MCAT had a score under 29? a) First step: Draw it! This time we can't find the proportion using the 68-95-99.7 rule, so we must calculate the Z-score. Recall, a z-score takes any normal distribution and \"standardizes\" it into a standard normal distribution, N(0,1), that we call \"Z\". X z score= Now, all we have to do is input values. Calculate a z score and look up the z score in a z table. The proportion that scores less than 29 is 0.736 (rounded). 2925 P ( X <29 )=P Z < =P ( Z <0.63 ) 6.4 ( ) Probability from a Z table 0.73565 Z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 .00 .50000 .53983 .57926 .61791 .65542 .69146 .72575 .75804 .78814 .81594 .01 .50399 .54380 .58317 .62172 .65910 .69497 .72907 .76115 .79103 .81859 . .50798 .54776 .58706 .62552 .66276 .69847 .73237 .76424 .79389 .82121 .03 .51197 .55172 .59095 .62930 .66640 .70194 .73565 .76730 .79673 .82381 .04 .51595 .55567 .59483 .63307 .67003 .70540 .73891 .77035 .79955 .82639 .05 .51994 .55962 .59871 .63683 .67364 .70884 .74215 .77337 .80234 .82894 . .52392 .56356 .60257 .64058 .67724 .71226 .74537 .77637 .80511 .83147 .07 .52790 .56749 .60642 .64431 .68082 .71566 .74857 .77935 .80785 .83398 .08 .53188 .57142 .61026 .64803 .68439 .71904 .75175 .78230 .81057 .83646 .09 .53586 .57535 .61409 .65173 .68793 .72240 .75490 .78524 .81327 .83891 b) Now use the TI-84 to find the proportion. Step 1: Go to 2nd Dist (above VARS). Choose Option 2:normalcdf( Step 2: Enter in the lower bound, upper bound, , for the proportion you want. Note: -1E99 is - , the lower bound. normalcdf(lower, upper, , ) = normalcdf(-1E99, 29, 25, 6.4). Press enter. 3 The proportion of students who take the MCAT and receive a score under 29 is 0.7340 or 73.40%. (Slight difference due to rounding) 4 8) Suppose to be admitted to Harvard Medical School a student must score above the 90th percentile on the MCAT. What is the lowest score a student can achieve to be eligible for admission? Note: The 90th percentile means that 90% of the data are below this value. So this question could also read, \"90% of the MCAT scores are below what value?\" a) First step: Draw it! we can't find the 90th percentile using the 68-95-99.7 rule, so we must use the Ztable. First look up the z score associated with .90 probability (or the closest thing to it) in the table. We get a z = 1.285. Which means P ( Z <1.285 ) 0.90 Again Z 1.0 1.1 1.2 1.3 1.4 .00 .84134 .86433 .88493 .90320 .91924 .01 .84375 .86650 .88686 .90490 .92073 . .84614 .86864 .88877 .90658 .92220 .03 .84849 .87076 .89065 .90824 .92364 .04 .85083 .87286 .89251 .90988 .92507 .05 .85314 .87493 .89435 .91149 .92647 . .85543 .87698 .89617 .91309 .92785 .07 .85769 .87900 .89796 .91466 .92922 .08 .85993 .88100 .89973 .91621 .93056 .085 .09 .86214 .88298 .90147 .91774 .93189 Now plug it into the formula below. Note this is just the z score calculation just \"rearranged\" to calculate a value x instead of a probability. x25 x25 P Z< 0.90 =1.285 x=1.285 6.4+ 25=33.22 6.4 6.4 ( ) Note: You can always use the formula x value=z + to get the x-value for a percentile. A person taking the MCAT in 2010 must score at least a 33.22 on the MCAT to get into Harvard Medical school. 2. Now use the TI-84. How high of a score does a student need to be in the top 10% of all students taking the MCAT? Step 1: Go to 2nd Dist (above VARS). Choose Option 3:invNorm( Step 2: Enter in the percentile, , . Note: Always enter in the proportion BELOW the value you need. invNorm(percentile, , ) = invNorm(0.90, 25, 6.4). Press enter. Part II: ST 201 Activity 3 The Normal Distribution Submit only pages from this section. 40 Points 5 Name______________________________________________ 1) (6 points) List at least three random variables that would most likely be normally distributed? Explain your reasoning. Examples: Height of a particular tree in a forest, Weights of babies born, Widths of pen barrels 2) (14 points) Emissions of sulfur dioxide by industry set off chemical changes in the atmosphere that result in acid rain. The acidity of liquids is measured by pH on a scale from 0 to 14. Distilled water has pH of 7.0 and lower pH values indicate acidity. The pH of rain at one location varies among rainy days according to a normal distribution with mean 5.41 and standard deviation 0.49. a. (4 points) Draw the normal distribution, label the mean, 1 , 2 , 3 . b. (2 points) Using the 68-95-99.7 rule. What is the approximate proportion of rainy days that have rain fall pH levels between 3.94 to 6.88? Suggestion: Try drawing the distribution and shading in the desired region. c. (2 points) Using the 68-95-99.7 rule. What is the approximate proportion of rainy days that have rain fall pH levels above 5.90? Suggestion: Try drawing the distribution and shading in the desired region. 6 d. (3 points) Normal rain is somewhat acidic, so acid rain is sometimes defined as rain fall with a pH below 5.0. What proportion of rainy days have rainfall with pH below 5.0? Calculate the probability using a z-score and the standard normal table or using TI-84. Show work or calculator command used. Suggestion: Try drawing the distribution and shading in the desired region. e. (3 points) What value represents the 25th percentile for rainy days? Show work or calculator command used. Suggestion: Try drawing the distribution and marking the approximate desired point. 3) (20 points) Graduate admissions at a particular university require applicants to take the GRE. Hong, June and Eduardo are three applicants to the same graduate program. Scores are valid for several years, in 2011 the GRE changed the scale of scores from 200-800 to 130-170. Each applicant took the exam during a different year. 7 Hong took the exam prior to the new scoring system, whereas June and Eduardo took the exam after the 2011 switch. The table below summarizes each of their scores: Applicants Score Year Average Standard Deviation Hong 458 2010 480 90 June 169 2011 154 8 Eduardo 160 2013 148 7 Applicant The distribution of GRE scores for each year is assumed to be normally distributed with the associated mean and standard deviation. a. (5 points) Why does it not make sense to compare the scores of the applicants in their given form? b. (6 points) Calculate a z score for each participant. Which of the students is the best applicant based solely on their GRE score? : Applicant Z score Hong June Eduardo 8 c. (6 points) For first consideration to the graduate school, a student's score must be at or above the 90th percentile. Which, if any, of the students made this cutoff? Calculate the percentile of each. Applicant Percentile Hong June Eduardo d. (3 points) Suppose in 2012 the middle 95% of GRE scores were from 135 to 165, assuming the distribution of scores is normal what is the mean and standard deviation? Hint: Draw the distribution and use the 68-95-99.7 Rule. 9