STP 231 EXAM #1 (Example) Instructor: Ela Jackiewicz Honor Statement: I have neither given nor received information regarding this exam, and I will not do so until all exams have been graded and returned. PRINTED NAME: Signed Date: DIRECTIONS: This is a closed book examination. You may use a graphing calculator and one 8x11 page with hand written notes, no completely solved problems are allowed. Turn in the notes with your exam. Formulas and tables are provided on the back of the test. There are 16 questions. For Questions#1 #6 provide complete and well organized answers showing all your work! Answers without supporting work will get no credit. Questions#7 #15 are multiple choice. Select appropriate letter answer (A to E as appropriate) and place your answer in the table provided below. Last problem is for Extra Credit, select True or False for each of 5 parts. Relax and good luck! Use table below to write letter answers (A E as appropriate) to questions 7 15, and circle True or False for Extra Credit Questions A E Question#7 Question#8 Question#9 Question#10 Question#11 Question#12 C C D C B B Question#13 Question#14 Question#15 A C E Extra Credit Questions: A) False B) True C) True D) False E) True
Use following information for questions #1 and #2 The IQ test scores for a random sample of 10 fifth grade students are: 96,, 118, 120, 122, 125, 126, 130, 135, 138 Question #1 ( 8 points) Give the 5 number summary of the data (MIN, MAX, Median, Q1 and Q3) and make a box plot. Use a line below to mark a proper scale. MIN= 96 Q1= 118 Median= 123.5 Q3= 130 MAX= 138 90 95 100 105 115 120 125 130 135 140 Question #2 (8 points) Check if there are any outliers in the data, by using IQR. List potential outliers, if there are no outliers, state it. Show all work, clearly justify your answer! IQR=12 1.5(12)=18 Lower Fence= 118 18=100 Upper fence=130+18=149 96 is below lower fence=potential outlier Question #3 (8 points) Use tables to compute 90 th percentile of N(5,2) curve. Clearly show work, include appropriate sketch. Sketch should represent a normal curve centered at 5, y=marked point above 5, You should show that area shaded left of y=90% (0.9) SKETCH: ANSWER: z=1.28, y=5+1.28(2)=7.56
Use following information in Questions #4 and #5 The distribution of scores for persons over 16 years of age on the Wechsler Adult Intelligence Scale (WAIS) is approximately normal with mean 100 and standard deviation 19. The WAIS is one of the most common IQ tests for adults. Question#4(8 points)what percentage of the persons over 16 have a WAIS score between 60 and 120? Include appropriate sketch illustrating your answer. Sketch should represent a normal curve centered at 100, y1=60 and y2=120 =marked points below and above 100. You should show that your answer=area shaded between 60 and 120 z1= 2.11 z2=1.05 0.8531 0.0174=0.8357 SKETCH: ANSWER: 83.57% Question#5(8 points) Suppose Y is a WAIS score of a randomly chosen individual over 16, what is the probability P(Y>130)? Include appropriate sketch illustration your answer. Sketch should represent a normal curve centered at 100, y=130 =marked point above 100. You should show that your answer=area shaded above 130. z=1.58 area=0.0571 SKETCH: ANSWER: 0.0571 Question#6 (6 points)based on a large sample, Average North American newborn is on average 20 inches long with standard deviation of 1.12 inch and has average weight of 7.5 pounds with standard deviation 1.25 pounds. Compute Coefficient of Variation for each measure and use it to answer the following question: Which of the two measures (length or weight) has greater variability? Coefficients of variation: Length 5.6% Weight: 16.7% Weight has greater variability
Questions #7 #15, 6 points each, are multiple choice, select a correct response and mark a letter of that response in the table given on the front page. Use the same table to mark T or F responses for Extra Credit Problem #16 Question#7 (6 points) The National Cancer Institute estimates that 3.65% of women in their 60 s get breast cancer. A mammogram can correctly identify 85% of cancer cases (positive test) and 95% of cases without cancer (negative test), which means we get false negative 15% of the time and false positive 5% of the time. Compute probability that randomly selected women in her 60 s will have a positive test for a breast cancer. Round to 2 decimal places A) 0.90 B) 0.85 C) 0.08 D) 0.03 E) none of these Use following information for questions #8 and #9 The probability model for response Y = In the past seven days, how many times did you go to a fitness center to work out? Y=Days 0 1 2 3 4 5 6 7 Probability.56.14.12.08.04.03.02.01 Question#8 ( 6 points) Compute the probability P(Y < 3 or Y > 5) A) 0.93 B) 0.96 C) 0.85 D) 0.15 E) none of these Question#9 ( 6 points) Compute mean (expected value ) of a random variable Y, round answer to 2 decimal places A)0.14 B) 1.68 C) 0.21 D) 1.12 E) none of these Use following data for questions #10 and #11 Consider the following stem & leaf diagram of the data it represents. There are 41 observations. 0 5 5 7 8 9 Data : Mercury Content (in the hair) of 1 0 2 3 5 7 7 8 8 9 Seychelles Fishermen 2 1 2 2 3 3 4 5 6 6 8 8 9 3 0 2 2 3 4 5 5 7 8 Stems: tens 4 2 5 6 5 8 leaves: ones 6 7 2 8 3 Question#10 ( 6 points) Which of the following best describes the shape of the distribution? Select one. A) Bell shape B) Uniform C) Right skewed D) Bimodal E) Symmetric Question#11 ( 6 points) Compute the median of this data. A) 5 B) 25 C) 27.37 D) 21 E) none of these
Use table below for questions #12 #13 The following table summarizes blood types and Rh types for typical people. Suppose one person is randomly selected. Compute following probabilities, round answers to 4 decimal places. Blood Type O A B AB total Rh positive 40 36 10 5 91 Rh negative 8 6 3 2 19 total 48 42 13 7 Question #12 ( 6 points) Compute probability that this person does not have blood type B A) 13 B) 97 C) 90 D) none of these Question #13 ( 6 points) Compute probability that that this person has blood type AB or is Rh positive. A) 93 B) 98 C) 5 D) none of these Use following information for questions #14, and #15 9% of men cannot distinguish between the colors red and green. If 8 men are randomly selected for a study of traffic signal perception, compute the following probabilities, round answers to 4 decimal places, if needed. Question #14 ( 6 points) Compute probability that exactly three of the men cannot distinguish between the colors red and green A) 0.09 B) 0.27 C) 0.0255 D) 0.9966 E) 0.0007 Question #15 ( 6 points) Compute probability that at least two of the men cannot distinguish between the colors red and green A) 0.18 B) 0.0081 C) 0.1289 D) 0.0289 E) 0.1577 Extra Credit Questions (5 points) Decide if statements below are true or false A) Sample standard deviation and sample mean are robust measures; they are resistant to outliers. B) Suppose test scores in a large sample of Mat 117 students have nearly bell shaped distribution with mean of 70 points and standard deviation of 10 points. According to Empirical Rule we can expect about 95% of all test scores to be between 50 and 90 points C) If we compare two histograms representing distribution of the same variable in the same scale, wider histogram will indicate higher variability. D) We do not expect any error when we estimate a population parameter from a sample statistics E)It is possible that median of a data set is not an actual observation.
n y i FORMULAS Sample statistics Sample mean:, Sample standard deviation (definition) y = n Sample variance= s 2 Range=Max Min s = n (y i y ) 2 n 1 Interquartile Range IQR= Q 3 Q 1, Lower Fence =Q1 1.5(IQR), Upper Fence=Q3+1.5(IQR) Coefficient of variation= s y 100% Population parameters: N Population mean: y i μ= N Standard score or z score z = y μ σ Population standard deviation: σ = N ( y i μ) 2 N Random Variables n μ Y = y i P (Y =y i )= y i p i μ Y =np for Binomial Y Probability 0 P E 1 P E c =1 P E P E 1 or E 2 =P E 1 P E 2 P E 1 and E 2 P E 1 and E 2 =P E 1 P E 2 Conditional probability P E 2 / E 1 = P E 1 and E 2 P E 1 for independent events only Binomial Distribution Formula n! P Y =k =C n,k p k 1 p n k, nck=c n,k = n k!k!