Answer:
The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Answer: ∴ The mean of new observation is 24 and Standard deviation of new observation = 12.
The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
Answer:
The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
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The variance of 20 observations is 4. If each observation is multiplied by 2, find the variance of the resulting observations.
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Find the mean, variance and standard deviation for the following data: 6, 7, 10, 12, 13, 4, 8, 12
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Find the mean, variance and standard deviation for the following data: 2, 4, 5, 6, 8, 17
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Compute the mean deviation from the median of the following distribution:
Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 10 20 5 10 Answer: Median is the middle term of the Xi, Middle term is 25 Median = 25 Class Interval xi fi Cumulative Frequency |di| = |xi – M|...
Find the mean deviation from the mean for the following data:
Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 Frequencies 4 8 9 10 7 5 4 3 Answer: = 17900/50 = 358 Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|...
Find the mean deviation from the mean for the following data:
Classes 95-105 105-115 115-125 125-135 135-145 145-155 Frequencies 9 13 16 26 30 12 Answer: = 13630/106 = 128.58 Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di|...
Compute mean deviation from mean of the following distribution:
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 8 10 15 25 20 18 9 5 Answer: = 5390/110 = 49 Class Interval xi fi Cumulative Frequency |di| = |xi – M| fi |di| 10-20 15 8...
The age distribution of 100 life-insurance policy holders is as follows:
Age (on nearest birthday) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5 No. of persons 5 16 12 26 14 12 6 5 Calculate the mean deviation from the median age. Answer: N = 96...
Find the mean deviation from the mean and from a median of the following distribution:
Marks 0-10 10-20 20-30 30-40 40-50 No. of students 5 8 15 16 6 Answer: N = 50 N/2 = 50/2 = 25 The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28...
Calculate the mean deviation from the median of the following frequency distribution:
Heights in inches 58 59 60 61 62 63 64 65 66 No. of students 15 20 32 35 35 22 20 10 8 Answer: Median is the Middle term, Median = 61 xi =Heights in inches fi = Number of students xi fi...
The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:
Number of calls 0 1 2 3 4 5 6 7 Frequency 14 21 25 43 51 40 39 12 Compute the mean deviation about the median. Answer: Median is the even term, (3+5)/2 = 4 Median = 8 xi =Number of calls fi =...
Calculate the mean deviation about the median of the following frequency distribution:
xi 5 7 9 11 13 15 17 fi 2 4 6 8 10 12 8 Answer: N = 50 Median = (50)/2 = 25 Median Corresponding to 25 is 13 xi fi Cumulative Frequency |di| = |xi – M| = |xi – 61| fi |di| 5 2 2 8 16 7 4 6 6...
Find the mean deviation from the mean for the following data:
xi 5 7 9 10 12 15 fi 8 6 2 2 2 6 Answer: xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di| 5 8 40 4 32 7 6 42 2 12 9 2 18 0 0 10 2 20 1 2 12 2 24 3 6 15 6 90 6 36 Total = 26 Total...
Find the mean deviation from the mean for the following data:
xi 5 10 15 20 25 fi 7 4 6 3 5 Answer: xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di| 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 Total = 25 Total = 350...
Find the mean deviation from the mean for the following data:
xi 10 30 50 70 90 fi 4 24 28 16 8 Answer: xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di| 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 Total =...
Find the mean deviation from the median for the following data :
xi 15 21 27 30 fi 3 5 6 7 Answer: N = 21 Median = (21)/2 = 10.5 Median Corresponding to 10.5 is 27 xi fi Cumulative Frequency |di| = |xi – M| fi |di| 15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30...
Find the mean deviation from the median for the following data :
xi 74 89 42 54 91 94 35 fi 20 12 2 4 5 3 4 Answer: N = 50 Median = (50)/2 = 25 Median Corresponding to 25 is 74 xi fi Cumulative Frequency |di| = |xi – M| fi |di| 74 20 4 39 156 89 12 6 32 64...
Find the mean deviation from the median for the following data :
Marks obtained 10 11 12 14 15 No. of students 2 3 8 3 4 Answer: N = 20 Median = (20)/2 = 10 Median Corresponding to 10 is 12 xi fi Cumulative Frequency |di| = |xi – M| fi |di| 10 2 2 2 4 11 3...
Calculate the mean deviation about the median of the following observation : (i) 3011, 2780, 3020, 2354, 3541, 4150, 5000 (ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
Answers: (i) Arrange the numbers in ascending order. Median is the middle number of all the observation. 2354, 2780, 3011, 3020, 3541, 4150, 5000 Median = 3020 and n = 7 Calculating Mean Deviation,...
Calculate the mean deviation about the median of the following observation : (i) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 (ii) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
Answers: (i) Arrange the numbers in ascending order. Median is the middle number of all the observation. 30, 34, 38, 40, 42, 44, 50, 51, 60, 66 Here the Number of observations are Even then Median =...
Calculate the mean deviation about the median of the following observation : 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Answer: Arrange the numbers in ascending order. Median is the middle number of all the observation. 34, 38, 43, 44, 47, 48, 53, 55, 63, 70 Here the Number of observations are Even then Median =...
Calculate the mean deviation from the mean for the following data : (i) 4, 7, 8, 9, 10, 12, 13, 17 (ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answers: (i) 4, 7, 8, 9, 10, 12, 13, 17 |di| = |xi – x| ‘x’ be the mean of the given observation. x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8 = 80/8 = 10 Number of observations, ‘n’ = 8 xi |di|...
Calculate the mean deviation from the mean for the following data : (i) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 (ii) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answers: (i) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 |di| = |xi – x| ‘x’ be the mean of the given observation. x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]/10 = 500/10 = 50 Number of...
Calculate the mean deviation from the mean for the following data : 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Answer: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59 |di| = |xi – x| ‘x’ be the mean of the given observation. x = [57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59]/10 = 550/10 = 55 Number of...
Calculate the mean deviation of the following income groups of five and seven members from their medians:
I Income in ₹ II Income in ₹ 4000 3800 4200 4000 4400 4200 4600 4400 4800 4600 4800 5800 Answer: Data is arranged in ascending order, 4000, 4200, 4400, 4600, 4800 Median = 4400 Total...
The lengths (in cm) of 10 rods in a shop are given below: 40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2 (i) Find the mean deviation from the median. (ii) Find the mean deviation from the mean also.
Answers: (i) Arrange the data in ascending order, 15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0 |di| = |xi – M| The number of observations are Even then Median = (40+52.3)/2 =...
In 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 find the number of observations lying between – M . D and + M . D , where M.D. is the mean deviation from the mean.
Answer: |di| = |xi – x| ‘x’ be the mean of the given observation. x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10 = 455/10 = 45.5 Number of observations, ‘n’ = 10 xi |di| = |xi –...
In 22, 24, 30, 27, 29, 31, 25, 28, 41, 42 find the number of observations lying between – M . D and + M . D , where M.D. is the mean deviation from the mean.
Answer: 22, 24, 30, 27, 29, 31, 25, 28, 41, 42 |di| = |xi – x| ‘x’ be the mean of the given observation. x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10 = 299/10 = 29.9 Number of...
In 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 find the number of observations lying between – M . D and + M . D , where M.D. is the mean deviation from the mean.
Answer: 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 We know that, |di| = |xi – x| ‘x’ be the mean of the given observation. x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]/10 = 494/10 = 49.4...
The following are some particulars of the distribution of weights of boys and girls in a class:
Which of the distributions is more variable? Solution:
An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:
(i) Which firm A or B pays out the larger amount as weekly wages? (ii) Which firm A or B has greater variability in individual wages? Solution:
Calculate coefficient of variation from the following data:
Solution:
The coefficient of variation of two distribution are 60% and 70%, and their standard deviations are 21 and 16 respectively. What is their arithmetic means?
Here, the Coefficient of variation for the first distribution is 60 And, Coefficient of variation for the first distribution is 70 $\mathrm{SD}\left(\sigma_{1}\right)=21$ and...
The means and standard deviations of heights and weights of 50 students in a class are as follows:
Which shows more variability, heights or weights? Solution: Given: The mean and SD is given of 50 students. Let us find which shows more variability, height and weight. By using the formulas,...
Two plants A and B of a factory show the following results about the number of workers and the wages paid to them
In which plant A or B is there greater variability in individual wages? Solution: Variation of the distribution of wages in plant $\mathrm{A}\left(\sigma^{2}=18\right)$ So, Standard deviation of the...
Calculate the mean, median and standard deviation of the following distribution
Solution:
A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure is 40. Find the correct mean and S.D.
Calculate the A.M. and S.D. for the following distribution:
Solution:
Calculate the standard deviation for the following data:
Solution:
Calculate the mean and S.D. for the following data:
Solution:
Find the standard deviation for the following data:
(i) Solution: (ii) Solution:
Find the mean, and standard deviation for the following data:
(i) Solution: (ii) Solution:
Table below shows the frequency f with which ‘x’ alpha particles were radiated from a diskette
Calculate the mean and variance. Solution:
Find the standard deviation for the following distribution:
Solution: By using the formula for standard deviation: $\begin{array}{l} \mathrm{SD}=\sqrt{\operatorname{Var}(\mathrm{X})} \\ \text { Mean }=\sum \frac{\mathrm{f}_{\mathrm{i}}...
7. The following table shows the ages of the patients admitted in a hospital during a year:
Ages (in years):5 – 1515 – 2525 – 3535 – 4545 – 5555 – 65No of students:6112123145 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. ...
10. Find the missing frequencies and the median for the following distribution if the mean is
Solution: It’s given that, \[N\text{ }=\text{ }200\] \[\Rightarrow 46\text{ }+\text{ }x\text{ }+\text{ }y\text{ }+\text{ }25\text{ }+\text{ }10\text{ }+\text{ }5\text{ }=\text{ }200\] \[\Rightarrow...
8. The following table gives the distribution of the life time of
neon lamps:
Find the median life. Solution: It’s seen that, the cumulative frequency just greater than \[n/2\text{ }\left( 400/2\text{ }=\text{ }200 \right)\text{ }is\text{ }216\]and it belongs to the class...
7. The following table gives the frequency distribution of married women by age at marriage.
Calculate the median and interpret the results. Solution: Here, we have \[N\text{ }=\text{ }357,\] So, \[N/2\text{ }=\text{ }357/\text{ }2\text{ }=\text{ }178.5\] The cumulative frequency just...