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Please answer the questions clearly. Source needed: PFIZER LTD (PFIZER.NS) Stock Historical Prices & Data - Yahoo Finance Directions: 1Write a report of your answers

Please answer the questions clearly. Source needed: PFIZER LTD (PFIZER.NS) Stock Historical Prices & Data - Yahoo Finance

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Directions: 1Write a report of your answers on separate sheets of paper, taking care to prepare a legible and organized report. Put your name on your report. Answer questions in order and underline or draw a box around any nal numerical answers. Show your work for all answers and include the company name you obtained the data from. Full credit will not be given to answers without work, and partial credit may be given when work is shown. Write any written answers or explanations clearly, and be as neat as possible. You will have to submit your assignment on Canvas as a mist tile (single le pleasel). You can embed pictures of your homework in a Word le and save it as pdf. Pro tip: use a scanning app such as \"CamScarmer.\" It's ee and it will produce quality pdf les. Notes on iPhones also a allows you to scan documents and save to pdf. 1. You will need the same dataset used for problem 3 in homework I (the dataset obtained 'om the yahoo website with the company you selected}. Use Excel to calculate the average and standard deviation of the close data column. Assume that these two numbers represent the population {=parametric} mean and population standard deviation, respectively, for the variable length {in cm) in a population of a species of sh. Attach a printout of the data to your homework and write down the ticker code ou it. 3. Calculate the probability of sampling at random a sh that is smaller in size than the value you would obtain by subtracting halfthe standard deviation from the average [a will be equal to: p. (o b. Calculate the probability of sampling at random a sh that is greater in size than the value you would obtain by adding halfthe standard deviation orn the average [x = p. + (#2)] c. Calculate the probability of sampling at random a sh that has a size between the two values [a = p ((#2), x = p. + (5152)] used in parts \"a" and "b,\" respectively {1. Calculate the 25th and T5111 percentiles of sh size for the population using the normal distribution table. e. Imagine that 5 individuals are sampled at random from this fish population. Calculate the probability that the average calculated will be less than the value: H - (6/3) NOTE: Assume the variable is normally distributed and use bell-shaped curve diagrams to shade the areas that correspond with the answers to questions "a" through "d"2. Write the last two digits of your l student ID number with a decimal point in between these two digits: Assume that this number is the average number of squirrels you come across when you walk from the Life Science Building {L5} to the Engineering Research Building (ERR). Tomorrow you plan to walk from L5 to ERB and you wonder what it the probability of observing at most three squirrels. Assume that squirrels are randomly distributed and that each squirrel is an independent observation. NOTE: if your last two digits happen to be both I}, use 4.2 as the average number of squirrels encountered. 3. Write the last digit of your 1000 student ID number: _ You know that in a specic population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a random sample of 9 individuals 'om this population: a. Calculate the probability that _ (last digit of your ID number} carry intestinal parasites. b. Calculate the probability that at least two individuals carry intestinal parasites. NOTE: you can still calculate \"a\" if your last digit is \"ll.\" PROBLEM 1: PROBLEM 2: No. of Relative Cumulative Relative Cumulative Calls Frequency Frequency Frequency Rewrite the given data vertically in a column of a table. Then compute for Frequency the sum of blood glucose levels Ex. deviation x - X. squared deviation (x - X). 3.333 3.333 and summation of the squared deviation E(x - X). 8.067 6.667 13.333 Blood Glucose Deviation Squared Deviation 13.333 20:807 Level (x) (x- 7)2 6.667 43.333 13.80 -0.47 0.22 23.333 66.667 ...15.10 0.83 0.69 20.000 86.667 13.70 -0.57 0.32 10.000 96.667 14.40 0.13 0.02 3.333 100.000 15.30 1.03 1.06 100.0 143.333 14,00 -0.27 0.07 12.30 -1.97 3.88 12.50 -1.77 3.13 15.50 1.23 1.51 Bar Graph of the Frequencies of Courtship Calls 16.10 3.35 Ex= 142.70 Xx - 1) = 14.26 The average or mean I, is calculated using the formula. Frequency 7- 142.70 2 3 5 = 14. 27 Number of Courtship Calls The data tell us that we have a sample of 10 biology students. We can use the formula below to calculate the standard deviation. The graph above compares the frequencies of the courtship calls made by Male purple dart-poison frogs during the afternoon. 5 is the mode or the most common number of courtship calls with 7 times, followed by 6. 4, 3. and others. "-1 The frequencies run from 1 to 8, giving a range of 7. Moreover, the frequencies look like a bell-shape, suggesting that the distribution of frequencies is normal among the number of courtship calls. 14.26 10 - 1 = 1.26 Therefore, the standard deviation of the blood glucose levels for ten biology students is 1.26 millimoles per liter. PROBLEM 3: Step S: For the Box Plot: Draw a horizontal number line that skips by 100's from I have selected the Pfizer Limited (PFIZER NS) available andperiod?_16093 5,100 to 5,400. Then, plot the 5-number summary. (My strategy: Begin http::/finance.yahoo.com/quote/PFIZER NS/history?period1=160673090 with the lowest, followed by highest, median, lower quartile, then 72800&interval=1d.& filter-history&frequency 1d.& includeAdjustedClose-true upper quartile) Date Open : High Low Close Ac Close Volume 54:020 8145 5154.05 5085 5107.85 5107.65 40414 5140 5385 SOPO : 5247 25 82:27 25 1049719 12/3/2020 5350 $436.6 $275 5318.65 5318 65 922316 12/4/2020 8259 9295.4 12/7/2020 5305.95 5386 95 83118 8937.5 9357 5 261712 S. 100 5,150 5.200 5. 5.300 5.350 1. 12/8/2020 5359 / 5300.05 : 5314.15 $31415 125293 1 12/9/2020 1 5424.21 5427.9 5281 8314.85 252589 12/10/2020 5209 5209| $100 2 5 $202 1 $202.1 172504 12/11/2020 | 8231.15 5920 5227 15201.15 9281.15 144904 12/14/2020 PROBLEM 4: 9310.3 8310.4 194059 12/15/2020 [ $300. 5310 $126 5256.5 $286.5 106788 When dealing with probability, it is a must that we know the sum of all 12/18/2020 5279 | 5210.1 5256.7 5254.7 12/17/2020 possible outcomes for each event. So, let's have the sum for each: 5285 5220 9214 06 12/18/2026 Fatal Not Fatal SUM 12/21 /2020 9415.19 $245 147936 No Seat Belt 1.563 161,220 162789 12/22/2020 $100 51802 51802 82606 Seat Bell 564 412.482 12/23/2020 5250 SUM 2.127 573.702 575.829 12/24/2020 4023.1 0437 95 8190 280459 12/28/2020 . sa30 $250 | $175 49 @) P(Not Worn) = m of No Tom Bell = 12ma =ANY 17ARI = 71317 = 0.2827 or 28.27% 12/29/2020 5202 5224.$ 5103 51532 $183.2 98749 b) P(Fatality Injury) = men areentering = = = 0.0037 or 0. 37%% -) P(Fatality Injury ( No Seat Belt) Sum of Fatal Injury Sum of No Seat Belt To compute for the five-number summary. 162.983 1959 Grand Total Step 1. List the data in ascending order, 575829 575,825 =0.0010 or 0.10%% 5.107.65 5,112.5 5,153.2 5,169.95 5,180.2 5.198.9 5,2021 [Or we can simply multiply what atained in a and b: 0.2827 0.0037) 5.203 5,212 8,225.5 5,227.25 5,246.85 5.259.4 5.256.5 d) P(Fatality Injury u No Seat Belt) 5.256.7 8.281.15 8.310.3 8.914.15 5.31485 5.918.65 5.937.5 when or rotel inmark , form of Me last fall - P(Fatality Injury ( No Seat Belt) 2.127 Step 2. Find the median, Position of Median = = 1: =11. The eleventh 575 29 + 375,823 -0.001 observation is the median, which is 5.227.2 = 0.2863 -0.0010 Step 2: Find the Lower Quartile, or Quartile 1. Position of Q1 = () = 21:1 = 5.5. = 0. 2854 or 28. 54% The observation between the 5" and 6' (or their average) is the Q1, that is ly exclusive jean ha 1802+31989 = 5,189.55 P(A UB) = PG) + P(B) - P(AnB) Step 3: Find the Upper Quartile, or Quartile 3, Position of (3 = "NL) = 121+1)= 16.5. e) P(Fatality Injury [No Seat Belt) The observation between the 18- and 17 is the Q3, that is 4281.15-83192 P(No Seat Belt n Fatality Injury) P(No Seat BOLF) 5, 295. 73. 0.0100 tep 4: Determine the lowest and es: 5,107.65 and 5,337.5 = 0. 0354 or 3.54% Source of the Data in Problem 3: https://finance.yahoo.com/quote/PFIZER.NS/history? period1=1606780800&period2=1609372800&interval=1d&filter=history&frequency=1d&includeAdjustedClose=true

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