1. Statistics for Entrepreneurs by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – PGPSE PARTICIPANTS mobile : 91+9414430763
6. What do you understand from Deseasonalisation of Data Removing the impact of seasonality is called deseasonalisation. Here we divide each element of data by seasonal index. Here we assume that the impact of seasonality is uniform throughout the data.
7. Example : Data for sale of Air conditioners is as under : summer 09 : 20000 winter 09: 1200 summer 2010 : 21000 seasonality index for summer is 400 and for winter is 20 thus the sale should be :20000/400 *100 = 5000 thus sale is : 5000, 6000, 5250 ...
8. What are the methods of forecasting ... Survey method / trend analysis methods / index number based methods survey can be of following types : 1. complete enumeration method 2. sample survey method in complete enumeration, we contact each person. It is also called census survey method.
9. What is delphi technique ? It is a teachnique to collect opinion of experts. Here you have to find out experts who are willing to give their projection. Then we obtain opinion of experts and try to compile their opinion. Each ideas is tested out and feasibility report is prepared. Thus final ideas is accepted on the basis of delhipi we are able to get the opinion of experts.
10. What are the smoothing methods used in statistical analysis ? In research we try to use smoothing method in order to remove vriations in data. There are mainly 2 methods for this: (i) Method of Moving Average, discussed earlier and (ii) Method of Exponential Smoothing smooth = without variations.
11. What are Barometric Technique or Lead or Lag Technique Lead and lag are those events which have close association with those variables which we want to study. Thus using statistical techniques, we can predict relation between lead / lag and the variable that we want to study. Thus we are able to search out a new variable (let us say x) which is a lead or a lag event to the variable Y that we want to predict.
12. What is the difference between laspeyre and paasche index? Laspeyre uses base year quantity paasche uses current year quantity as weights. Thus this is the only difference.
13. Do you agree : Arithmetic mean can be calculated for a distribution with open ends. Distribution means a range of data. Distribution with open end means – there is no opening no closing limit to the distribution. No – unless it is specified that it is normal distribution
14. The mean of a certain number of observations is 40. If two or more items with values 50 and 64 are added to this data, the mean rises to 42. Find the number of items in the original data. 40X + 50+64 = 42 (X+2) 40X-42X = -114+84 -2X = - 30 X = 15 answer
15. The mean salary paid per week to 1000 employees of an establishment was found to be Rs. 900. Later on, it was discovered that the salaries of two employees were wrongly recorded as Rs. 750 and 365 instead of 570 and 635. Find the corrected mean salary. Difference is : subtract (750-570) add (365-635) =270-180 = +90 divide 90/1000 = .09 so answer is : 900+.9 = 900.09
16. Find median of the observations in each of the following two cases : A : 25, 14, 28, 30, 25, 15, 32 B: 35, 20, 55, 27, 15, 40 Step 1. : arrange them Step 2 find mid position : 14,15,25,25,28,30,32 so median = 25 B : 15,20,27,35,40,55 so mid point is between 27 and 35, which is : (27+35)/2 = 31
17. Find the median of the distribution. Marks obtained 0-15 15-30 30-45 45-60 60-75 75-90 90-100 No. of students 26 34 64 76 60 30 10 Calculate CF : 26,60,124,200,260,290,300 the mid point is 150, which is in 45-60 range. =45 + (150-124) * 15/76 = 50.13 answer
18. Find median of the following data Age greater than (in yrs.) 0 10 20 30 40 50 60 70 230 218 200 165 123 73 28 8 Divide 230/2 = 115 (the values are in cumulative frequency) thus it is : between 40 to 50. =40+(123-115)*10/50 =40+1.6 = 41.6 answer
19. If the median of the distribution is Rs. 59.25, find the missing frequencies (total = 900) Wages (Rs.) 30-40 40-50 50-60 60-70 70-80 No. of Workers 120 ? 200 ? 185 First missing value is X 50+ (450-(120+x))*10/200 = 59.25 330-x = 185 -X =-145 or X = 145 total of all known values is 650 second value is 900-650 = 250 answer. :
20. Find mode in the above table : by observation it is 13 (highest frequency is 100 so this is mode), but if you prepare telly bars, the answer will be different. Since 13 is not the mid point and frequencies after this are constantly high, and frequencies below it are low, so we have to use telly bars to find exact mode. Based on this the answer is 14
22. Solution Formula : L1 + (f1-f0) / ((f1-0)+(f1-f2) * interval L1= lower limit of the modal class F1 = the value of highest frequency F0 = just before the modal class F2=just after the modal class =45+ (18-12) / ((18-12) +(18-14)) * 5 =45+6/10 * 5 = 45+3 = 48 answer
23. Mode and median are 75 and 60 what is mean? Mean – mode = 3 (mean – median) X-75= 3(X-60) X-3X = -180+75 -2X = -105 X = 52.5 answer
25. Solution 1. first of all calculate mean and median so that we can solve this problem. 2 then find difference of each value from the mean / median 3. square the values and total them up
27. Find mean deviation ? X : 10 11 12 13 F : 3 12 18 12 find mean, find its difference, multily difference with F and find the average : (30+132+216+156)/45 =547/45 =12.16 2.16, 1.16, .16,.84 multiply these to frequencies : (6.48+13.92+2.89+10.08)/45= .74 ans
28.
29. Price of a commodity went up by 6%, 10% and 15% respectively, in the last three years. What is the annual average change of price? Let us assume the price was 100, it went up to 106 after 1 year and (106)*110/100 = 116.6, now it went up to : 116 * 115/100 = 134.9 the average growth is : (1.349)^/1/3 = 1.1 thus 10% approximately OR first multiply 1.06*1.1*1.15 and find cubic root (as there are 3 values) of the resulting number.
30. Calculate the range and its coefficient from the following data : 159, 165, 140, 125, 110, 170, 132, 150 Range = (highest – least value) =170-110= 60 Coefficient of range = 60 / (170+110) =60/280 * 100 = 21.4%
31. There are 60 male and 40 female workers in a factory. The standard deviation of their wages (per hour) were computed as Rs. 8 and 11 respectively. The mean wages of the two groups were found to be equal. Compute the combined standard deviation of the Wages of all the workers. Combined mean= (60*X+40X)/100 = X =sqrt((60*64)+(40*121)/100) =sqrt((3840+4840)/100)=sqrt(86.8) =9.31 answer (mean of both groups is same – so there is no need to calculate combined mean and the value is zero in formula).
32. The mean, standard deviation and range of a symmetrical distribution of weights of a group of 20 boys are 40, 5 and 6 Kgs. respectively. Find the mean and standard deviation of the group if the lightest and the heaviest boys are excluded. Highest value : 40+3 = 43, least value = 37 (assuming rage to be eqully spread). Mean will remain same. old variance : 25 Total of variance = 25*20 = 500 less variance of excluded value : (9+9) = 482 482/18 =26.78, its square root = 5.17, so new standard deviation is 5.17 ans.
33. For a group of 30 male workers, the mean and standard deviation of weekly overtime work (number of hours) are 10 and 4 respectively. For another group 20 female workers, the mean and standard deviation of weekly overtime work are 5 and 3 respectively. Calculate combined mean & Std. Deviation. Combined mean = ((30*10)(20*5)/(30+20)) =8 combined variance= ((30*16)+(20*9)/50 + (30*4)+(20*9)/50) ) =19.2 combined st.dev.= 4.38 answer
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