Can you please check my work and explain any wrong answers? Thank you and happy holidays!

Problems need to include all required steps and answer(s) for full credit. All answers need to be reduced to lowest terms where possible. If the answer is in %, show two decimal places. Answer the following problems showing your work and explaining (or analyzing) your results. Submit your work in a typed Microsoft Word document. Below are excel functions, which you can use to solve for measures of central tendency and variability. You can use the formulas too; but, they will be time consuming EXCEL FUNCTIONS Measures of Central Tendency Suppose data are in cells A1 to A10 Mean =AVERAGE(A1:A10) Median =MEDIAN(A1:A10) Mode =MODE(A1:A10) Measures of Variability Suppose data are in cells A1 to A10 Range =MAX(A1:A10)-MIN(A1:A10) IQR =QUARTILE.EXC(Array,3)-QUARTILE.EXC(Array, 1) Population Variance 02 =VAR.P(A1:A10) Sample Variance s2 =VAR(A1:A10) or VAR. S(A1:A10) Population Standard Deviation o =STDEV.P(A1:A10) Sample Standard Deviation s =STDEV(A1:A10) or STDEV.S(A1:A10)1. Joe and Mary would like to buy their first home in a new city. Below is the list of prices of 20 homes for sale. Prices in 140 190 265 115 270 240 250 160 201 U.S. Dollars 240 280 175 200 310 195 320 105 385 264 a. Find the mean, median, mode. (2 pts) b. Which measure of central tendency best represent the data? Why? (2 pts) C. Are there any outliers? If so, name the outlier. (2 pts) 2. The ages of students in a statistics class are listed below. Age of students 22 20 25 23 27 30 18 31 19 45 45 38 27 19 28 33 40 35 24 a. Create a stem plot (2 pts) b. What is the mean age of the students? (2 pts) 3. Jenny likes to have at least a 70% average to pass her statistics class. Her previous four test scores are 52%, 66%, 76% and 71%. What is the minimum score Heidi needs on the final exam to pass the class? (4 pts) 4. Consider the following data and corresponding weights. Weight (w ) 31 2.3 3 2.8 2 4 a. Compute the weighted mean. (2 pts) b. Compute the mean without weighting. (2 pts) 5. The chart below represents the amount of rainfall for 15 days in northern Ohio in July. Rainfall in inches 4.1 4.2 3.5 3.9 4.2 4.4 3.8 3.7 3.3 3.4 4.3 3.2 4.6 4.5 4.0 a. Complete the frequency table. (2 pts) Rainfall in inches Frequency Less than 3.0 3.0-3.4 3.5-3.9 4.0-4.5 4.8-5.0b. Find the mean of the raw data. (2 pts) c. Find the median of the raw data. (2 pts) d. Find the mode of the raw data. (2 pts) 6. Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28 and 25. a. What is the minimum? (2 pts) b. What is the first quartile? (2 pts) c. What is the median? (2 pts) d. What is the third quartile? (2 pts) e. What is the maximum? (2 pts) 7. Create the box plot for the data in #6. Label the five points on the box. (4 pts) 8. Using the data in Question 6: a. What is the range? (2 pts) b. What is the interquartile range (IQR)? (2 pts) 9. The histogram of a quantitative variable is positively skewed. The mean of the variable is 35. a. Which one of the following is a more likely value of the median? (2 pts) 55 30 35 b. Support your answer. (2 pts) 10. A histogram of a given data shows three clear peaks. This may mean that (4 pts) a. There are three distinct groups in the data b. There are three modes in the data c. Both a and b are correctMAT201 1. I have rearranged the numbers to make it easier to deal with 105, 115, 140, 160, 175, 180, 190, 195, 200, 201, 240, 240, 250, 264, 265, 270, 280, 310, 320, 385 A. mean = 224.25 median = 220.5 mode =240 B. The median is the best measure of central tendency in this case due to mean and mode being skewed by the outlier of 385 and the only duplicate number of 240. C. There are no outliers as our IQR is 90 and no values fall outside the upper and lower fence. 2. Order of number from least to greatest 18, 19, 19, 19, 20, 22, 23, 24, 25, 26, 27, 27, 30, 31, 33, 35, 36, 40, 45, 45 Table 1 Stem Plot Stem Leaf 8,9,9,9 0,2,3,4,5.6.7,7 AWNE 0,1,3,5,6 0,5.5 B. The mean age of the students is 28.2 which is found by adding all the data points which comes to 364 and dividing it by the number of points, 20. 3. 52%, 66%, 71%, 76%. With this set of data, we can find her current grade by finding the mean which is 66.25% in order to get a 70% she would need to raise it by 3.75%. we can determine the needed test score with the formula 70(goal mean)= 265+x(sum of data+ the final test score)/5(thenumber of points in the data including the final test). In this equation x=85 therefore Jenny would need to score an 85% in order to reach her goal of 70% overall grade. 4. organize the data and their corresponding weight 2.8, 2 2.3, 3 4, 4 3.1, 5 A. The formula for figuring out the weighted mean is 2.8(2)+2.3(3)+4(4)+3.1(5)/2+3+4+5 =Xw simplified (5.6+6.9+16+15.5)/14= 3.143 is the weighted mean B. The mean without weighting is (3.1+2.3+2.8+4)/4 or 3.05 5. 3.2, 3.3, 3.4, 3.5, 3.7, 3.8, 3.9, 4.0, 4.1, 4.2, 4.2, 4.3, 4.4, 4.5, 4.6 Table 2 A. Frequency Table Rainfall in inches Frequency Less than 3.0 3.0-3.4 3.5-3.9 4.0-4.5 4.6-5.0 B. Mean = 3.94 (59.1/15) C. Median = 4 (center point of the dataset) D. Mode = 4.2 (occurs 2x) 6. 15, 20, 25, 25, 27, 28, 30, 34\f8. a. Range = 19 = max-min = 34-15 b. IQR = 6.5 = Q3-Q1=29-22.5 9. a. 30 b. If the mean of the variable is 35 and positively skewed that would mean the majority of the numbers are on the lower end of the table and could not be higher than the median. 10. C , If there are 3 distinct peaks in the histogram then there could be 3 distinct groups in the data and or three modes in the data