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DataOff-Course Income ($1000s)10,5509,5009,2509,2508,1008,0007,650.docx
- 1. DataOff-Course Income ($1000s)10,5509,5009,2509,2508,1008,0007,6505,5505,3505,07 04,8254,7504,7504,6004,2754,0504,0004,0003,7253,7003,6753, 2753,2253,2003,1802,8502,5752,5002,4752,4502,4252,3001,92 51,9001,8751,7251,6501,6001,5001,350 Sheet2 Sheet3 Q1. If 2% of all high school graduates in the USA are admitted in Harvard university , and if 300,000 students are expected to graduate from high school this year. The mean and the standard deviation of the number of students that will be admitted to Harvard university next year? The probability of admitting 5000 or fewer students in Harvard university next year? Q2. A random sample of 40 smoking people is classified in the following table: Age Frequency 10 - 20 4 20 - 30 6 30 - 40 12 40 - 50 10 50 - 60 8 The mean age of this group of people ? Q3. In measuring the center of the data from a skewed distribution, the median would be preferred over the mean for most purposes because: (choose an answer) a.the median is the most frequent number while the mean is
- 2. most likely b.the mean may be too heavily influenced by the larger observations and this gives too high an indication of the center c.the median is less than the mean and smaller numbers are always appropriate for the center d.the mean measures the spread in the data e.the median measures the arithmetic average of the data excluding outliers. Q4. Suppose a random sample of 100 12-year-old boys were chosen and the heights of these 100 boys recorded. The sample mean height is 64 inches, and the sample standard deviation is 5 inches. You may assume heights of 12-year-old boys are normally distributed. Which interval below includes approximately 95% of the heights of 12-year-old boys? (choose an answer) a. 63 to 65 inches. b. 39 to 89 inches. c. 54 to 74 inches. d. 59 to 69 inches. e. Cannot be determined from the information given. f. Can be determined from the information given, but none of the above choices is Q5. Suppose that the age among elderly is distributed as follows: age Male Female 65 – 69 5 10 70 – 74 9 17 75 – 79 11 18
- 3. 80 - 84 8 12 85 + 4 6 The best display for showing the relationship between two categorical variables is the: (MCQ) A) frequency polygon. B) boxplot. C) scatterplot. D) contingency table. Q6. The standard normal probability distribution: (choose an answer) a. is a special case of the normal probability distribution. b. has a mean equal to 0 and a standard deviation equal to 1. c. measures the distance from the mean in units of the standard d. all of the above Q7. A normal probability distribution is completely described by (choose an answer) a. its mean. b. its standard deviation. c. its mean and standard deviation. d. none of the above. DataManagerAnnual SalaryTraining Program155769.50No250823.00Yes348408.20No449787.50No5 52801.60Yes651767.70No758346.60Yes846670.20No950246.80 Yes1051255.00No1152546.60No1249512.50Yes1351753.00Yes 1453547.10No1548052.20No1644652.50Yes1751764.90Yes184 5187.80Yes1949867.50Yes2053706.30Yes2152039.50Yes22529 73.60No2353372.50No2454592.00Yes2555738.10Yes2652975.1
- 22. 6859522.50No246954691.10No247057588.60Yes247152441.80 Yes247251030.10No247352840.40No247447996.80No24755014 3.20Yes247652209.30No247749576.70Yes247851015.70No247 958221.30Yes248053497.40Yes248151859.80Yes248250353.20 Yes248351847.30Yes248448110.20Yes248544947.10No248647 605.50Yes248752928.60Yes248850766.00Yes248951425.10Yes 249047295.80Yes249149369.70Yes249246155.90No249349045. 40No249450795.50No249550282.50Yes249656873.50No249747 860.70Yes249856543.70Yes249953168.80Yes250048433.30Yes PSTAT 109: Lab 6 Fall 2015 Instructions: When you finish all problems, ask your TA to check your work. Problem 1 (Comparing random samples and populations). Open the Excel file EAI.xlsx. The data show the annual salary for all 2, 500 managers in one company. a. What is the population mean annual salary? b. What is the population standard deviation for annual salary? Now, we want to construct 10 random samples each of size 25. One way to do this is as follows: (i) Assign a random number to each salary, so in the column adjacent to the salary column, use the RAND function generate a list of random numbers uniformly dis- tributed between 0 and 1.
- 23. (ii) Sort salaries by their corresponding random number. This can be accom- plished by using the sort feature found in the Data tab. Sort by either ascending or descending. (iii) Take the first 25 sorted salaries as our first random sample. Copy and paste the values of these 25 salaries into a new column. (iv) Press F9. This will reassign a new random number to each salary. (v) Repeat steps (ii), (iii), and (iv) until you have 10 random samples of size 25. For each of the 10 random samples, c. Find a point estimate of the population mean annual salary. (Consider using the fill-handle here.) d. Find a point estimate of the population standard deviation. (Consider using the fill- handle here.) e. How do the 10 point estimates of the population mean compare to the actual popula- tion mean? f. How do the 10 point estimates of the population standard deviation compare to the actual population standard deviation.
- 24. Problem 2. Open the Excel file GolfIncome.xlsx. The data represent income (in thou- sands of dollars) earned by a population of 40 professional golfers from their endorsement deals. a. What is the probability that a golfer earned between 4, 000 and 6, 000 thousand dollars? Hint: One way to find the number of incomes between 4, 000 and 6, 000 thousand dollars is to use the COUNTIFS function. b. How much did a golfer have to earn to be in the top 25% of the income bracket? Hint: Sort ascending the dataset, then find the (100 − 25) = 75th percentile (Recall the percentile formula back in lecture 2). Now, assume that the data follow a normal distribution. Using the NORM.DIST and NORM.INV functions, determine the following: c. What is the probability that a golfer earned between 4, 000 and 6, 000 thousand dollars? d. How much did a golfer have to earn to be in the top 25% of the income bracket?