Nm Shah 2018 Solutions for Class 11 Commerce Economics Chapter 6 Positional Average And Partition Values are provided here with simple step-by-step explanations. These solutions for Positional Average And Partition Values are extremely popular among Class 11 Commerce students for Economics Positional Average And Partition Values Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Nm Shah 2018 Book of Class 11 Commerce Economics Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Nm Shah 2018 Solutions. All Nm Shah 2018 Solutions for class Class 11 Commerce Economics are prepared by experts and are 100% accurate.

#### Question 1:

Calculate median of the following data:

 145 130 200 210 198 234 159 160 178 257 260 300 345 360 390

Arranging the data in ascending order.

130, 145, 159, 160, 178, 198, 200, 210, 234, 257, 260, 300, 345, 360, 390

#### Question 2:

Find out median of the following information:

 Marks : 10, 70, 50, 20, 95, 55, 42, 60, 48, 80

Arranging the data in ascending order.

10, 20, 42, 48, 50, 55, 60, 70, 80, 95

Here, the number of observations is even.

Thus, median marks is 52.5.

#### Question 3:

We have the following frequency distribution of the size of 51 households . Calculate the arithmetic mean and the median.

 Size 2 3 4 5 6 7 Total Number of households 2 3 9 21 11 5 51

 Size (X) Households (f) fX c.f 2 2 4 2 3 3 9 5 4 9 36 14 5 21 105 35 6 11 66 46 7 5 35 51 Æ©f = 51 Æ©fx = 255

Now, we need to look at the column of cumulative frequency. The item just exceeding the 26thitem is 35, which corresponds to 5.
Hence, the median is 5.

#### Question 4:

Find out median:
(a)

 Serial No. 1 2 3 4 5 6 7 8 9 Values 2 4 10 8 15 20 12 25 30

(b)
 X 5 10 15 20 25 f 2 4 10 8 15

(a) For the calculation of median the given values are first arranged in ascending order as follows.

 Serial Number Values Ascending Order 1 2 3 4 5 6 7 8 9 2 4 10 8 15 20 12 25 30 2 4 8 10 12 15 20 25 30
Here, N = 9

Thus, median is 12.

(b)
 X f c.f 5 2 2 10 4 6 15 6 12 20 8 20 25 10 30

Now, we need to look at the column of cumulative frequency. The item just exceeding the 15.5th item is 20, which corresponds to 20.
Thus, median is 20.

#### Question 5:

Find out median , first quartile and third quartile of the following series:

 Height (in inches) 58 59 60 61 62 63 64 65 66 No of Persons 2 3 6 15 10 5 4 3 1

 Height (X) Persons (f) c.f 58 2 2 59 3 5 60 6 11 61 15 26 62 10 36 63 5 41 64 4 45 65 3 48 66 1 49

(i)Median

This corresponds to height 61. Thus, the median is 61.

(ii) First Quartile

This corresponds to height 61. Thus, first quartile is 61.

(iii) Third Quartile

This corresponds to height 63. Thus, third quartile is 63.

#### Question 6:

The marks obtained by 68 students in an examination are given below. Compute the median.

 Marks Below 20 20-40 40-60 60-80 Above 80 No of students 0 5 22 25 16

 Marks Students (f) c.f Below  20 0 0 20 – 40 5 5 40 – 60 22 27 60 –80 25 52 Above 80 16 68

Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{68}{2}\right)}^{\mathrm{th}}$ item, which is the 34th item. This corresponds to the class interval 60−80, so this is the median class.

Thus, median is 65.6.

#### Question 7:

Calcualte the mean of the following distribution of daily wages of workers in a factory:

 Daily Wages (in â‚¹) : 100-120 140-160 160-180 180-200 Total No of Workers : 10 30 15 5 80

Also, calculate the median for the distribution of wages given above.

 Wages Worker (f) X fX c.f 100 – 120 10 110 1100 10 120-140 20 130 2600 30 140 – 160 30 150 4500 60 160 – 180 15 170 2550 75 180 – 200 5 190 950 80 80 Æ© fX = 11700

Note: In the above calculation we have assumed a class interval (120-140) and its corresponding frequency as 20.

#### Question 8:

The following table gives the marks obtained by 65 students in statistics in a certain examination . Calculate the median.

 Marks No of Students More than 70 8 More than 60 18 More than 50 40 More than 40 45 More than 30 50 More than 20 63 More than 10 65

 Marks Studentsc.f More than 70 More than 60 More than 50 More than 40 More than 30 More than 20 More than 10 8 18 40 45 50 63 65

Now the more than cumulative frequency distribution can be converted into class intervals as follows.

 C.I. f c.f 10 – 20 2 2 20 – 30 13 15 30 – 40 5 20 40 – 50 5 25 50 – 60 22 47 60 – 70 10 57 70 and above 8 65 Æ©f = 65

Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{65}{2}\right)}^{\mathrm{th}}$ item, which is the 32.5th item. This corresponds to the class interval 50−60, so this is the median class.

Thus, the median marks is 53.4.

#### Question 9:

Calculate the Arithmetic mean and median of the following frequency distribution:

 Class Interval 10-20 20-40 40-70 70-120 120-200 Total Frequency 4 10 26 8 2 50

 Class Interval f X (Midpoint) fX c.f 10 – 20 4 15 60 4 20 – 40 10 30 300 14 40 – 70 26 55 1430 40 70 – 120 8 95 760 48 120 – 200 2 160 320 50 Æ©f = 50 Æ©fX = 2870

$\overline{X}=\frac{\mathrm{\Sigma }fX}{\Sigma f}=\frac{2870}{50}=57.4\phantom{\rule{0ex}{0ex}}$
Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{50}{2}\right)}^{\mathrm{th}}$ item, which is the 25th item. This corresponds to the class interval 40−70, so this is the median class.

Thus, mean and median are 57.4 and 52.69 respectively.

#### Question 10:

An analysis for more efficiency in a factory , indicating the distribution of ages of workers was as follows:

 Age (in years) 16-19 20-29 30-39 40-49 50-59 60-64 Frequency 15 46 49 32 28 14
(a) Calculate the mean and median of the above data.
(b) Draw a histogram and indicate mode therein.

(a)

 Age Frequency (f) Midpoint  (m) fm 16 – 19 15 17.5 262.5 20 – 29 46 24.5 1127 30 – 39 49 34.5 1690.5 40 – 49 32 44.5 1424 50 – 59 28 54.5 1526 60 – 64 14 62 868 $\sum f=184$ $\sum fm=6898$

Mean

Hence, the mean of the given distribution is 37.48

Median

Note that the given distribution is in the form of inclusive class intervals. For the calculation of median, first the class intervals must be converted into exclusive form using the following formula.
Value of lower limit of one class  Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.
 Age Frequency (f) Cumulative Frequency(c.f) 15.5 – 19.5 19.5 – 29.5L1→29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 64.5 15 46         49→ f1 32 28 14 15          61→c.f 110 142 170 184

Median class is given by the
Size of
${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item = ${\left(\frac{184}{2}\right)}^{\mathrm{th}}$ item = 92th item.
This corresponds to the class interval of (29.5 39.5), so this is the median class.

Hence, the median is 35.826

(b)
Here the data is in the form of unequal class interval. So, we will first make appropriate adjustment in the frequencies to make the class intervals equal.
 Age Frequency (f) Adjusted Frequency 15.5 – 19.5 19.5 – 29.5 29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 64.5 15 46 49 32 28 14 $\frac{15×10}{4}=37.5$$-$$-$$-$$-$$\frac{14×10}{5}=28$

From the above graph, it can be seen that mode is 30.5. However, mean can not be determined graphically.

#### Question 11:

The marks obtained by 100 students  in an examination are given below. Compute the median, 1st and 3rd Quartiles.

 Marks 30-35 35-40 40-45 45-50 50-55 55-60 60-65 No of students 14 16 18 23 18 8 3

 Marks f c.f 30 – 35 14 14 35 – 40 16 30 40 – 45 18 48 45 – 50 23 71 50 – 55 18 89 55 – 60 8 97 60 – 65 3 100

N=100
Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is the 50th item. This corresponds to the class interval 45−50, so this is the median class.

First quartile class is given by the size of the ${\left(\frac{N}{4}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{100}{4}\right)}^{\mathrm{th}}$ item, which is the 25th item. This corresponds to the class interval 35−40, so this is the first quartile class.

Third quartile class is given by the size of the $3{\left(\frac{N}{4}\right)}^{\mathrm{th}}$ item, i.e. the $3{\left(\frac{100}{4}\right)}^{\mathrm{th}}$ item, which is the 75th item. This corresponds to the class interval 50−55, so this is the third quartile class.

#### Question 12:

Compute mode  from the following series:

 Size  of items : 2 3 4 5 6 7 Frequency : 3 8 10 12 16 14 Size of items : 8 9 10 11 12 13 Frequency : 10 8 17 5 4 1

 Size of Items Frequency 2 3 3 8 4 10 5 12 6 16 7 14 8 10 9 8 10 17 → Modal class 11 5 12 4 13 1

Here, mode is equal to 10 as it has the highest frequency.

#### Question 13:

Calcualte mode for the following data:

 No. of persons 1 2 3 4 5 6 7 8 9 10 Families 26 113 120 95 60 42 21 14 5 4

 No. of Persons (X) Families (f) 1 26 2 113 3 120 $\underset{}{\to }$Modal class 4 95 5 60 6 42 7 21 8 14 9 5 10 4

Here, mode is equal to 3 as it has the highest frequency.

#### Question 14:

Find out the Mode from any of the following two distributions:
(a)

 X : 30-40 40-50 50-60 60-70 70-80 80-90 90-100 f : 6 10 16 14 10 5 2

(b)
 Marks : 0-9 10-19 20-29 30-39 40-49 No of candidates : 6 29 87 181 247 Marks : 50-59 60-69 70-79 80-89 90-99 No of candidates : 263 113 49 9 2

(a)

 X f 30 – 40 6 40 – 50 10 50 – 60 16 $\underset{}{\to }$Modal class 60 – 70 14 70 – 80 10 80 – 90 5 90 – 100 2
By inspection, we can spot that (50–60) has the highest frequency, i.e. 16. So, this is the modal class.

Thus, the mode is 57.5.

(b)
Since this is an inclusive class interval series, to calculate mode, first we need to convert the inclusive class intervals into the exclusive class intervals by using the following formula.

The value of adjustment (as calculated) is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

 Marks f –0.5 – 9.5 6 9.5 – 19.5 29 19.5 – 29.5 87 29.5 – 39.5 181 39.5 – 49.5 247 49.5 – 59.5 263 $\underset{}{\to }$Modal class 59.5 – 69.5 113 69.5 – 79.5 49 79.5 – 89.5 9 89.5 – 99.5 2

By inspection, we can spot that (49.5 – 59.5) has the highest frequency, i.e.263. So, this is the modal class.

Thus, the mode is 50.46.

#### Question 15:

Life of electric lamps is given in the following table. Calculate the median and the mode.

 Life in  hours Number of lamps Below 400 4 400-800 12 800-1200 40 1200-1600 41 1600-2000 27 2000-2400 13 2400-2800 9 Above 2800 4

 Class Interval f c.f. Below 400 4 4 400 – 800 12 16 800 – 1200 40 56 1200 – 1600 41 97 1600 – 2000 27 124 2000 – 2400 13 137 2400 – 2800 9 146 above 2800 4 150

Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{150}{2}\right)}^{\mathrm{th}}$ item, which is the 75th item. This corresponds to the class interval 1200−1600, so this is the median class.

By inspection, we can spot that (1200–1600) has the highest frequency, i.e. 41. So, this is the modal class.

#### Question 16:

Calculate Mean, Median and Mode fromt he following data:

 Marks (mid points) No of Students 59 1 61 2 63 9 65 48 67 131 69 102 71 40 73 17 Total = 350

Converting the series into exclusive series

 C.I. m f. fm c.f 58 – 60 59 1 59 1 60 – 62 61 2 122 3 62 – 64 63 9 567 12 64 – 66 65 48 3120 60 66 – 68 67 131 8777 191 68 – 70 69 102 7038 293 70 – 72 71 40 2840 333 72 – 74 73 17 1241 350 Æ©f = 350 Æ©fm = 23764

#### Question 17:

Following is the distribution of marks of 50 students in a class:

 Marks (more than) : 0 10 20 30 40 50 No of students : 50 46 40 20 10 3
Calculate the Median Marks . If 60% of students pass this examination , find out the minimum marks obtained by a pass candidate.

Convert the series into exclusive class interval

 C.I. more than c.f. f less than c.f 0 – 10 50 4 4 10 – 20 46 6 10 20 – 30 40 20 30 → Median class 30 – 40 20 10 40 40 – 50 10 7 47 50 – 60 3 3 50

N = 50

If 60% students pass, it means 40% students fail. Thus, we have to find D4.
4thdecile class is given by the size of i.e. 20th item which falls in 30 cumulative frequency group.

#### Question 18:

The age in completed years of 50 persons is given below:

 32 61 52 56 22 49 97 35 30 30 95 67 42 20 31 64 20 10 62 60 27 53 31 90 54 25 43 47 35 21 43 75 45 22 36 13 46 23 51 11 15 39 50 42 77 73 81 40 40 55

Prepare frequency table taking 10-19, 20-29, ,.......... class intervals and calculate modal age.

 Age Tall marks No. of persons (f) Exclusive Class Interval 10 – 19 4 9.5 – 19.5 20 – 29 8 19.5 – 29.5 30 – 39 9 29.5 – 39.5 40 – 49 10 39.5 – 49.5 $\underset{}{\to }$Modal class 50 – 59 7 49.5 – 59.5 60 – 69 5 59.5 – 69.5 70 – 79 3 69.5 – 79.5 80 – 89 1 79.5 – 89.5 90 – 99 2 89.5 – 99.5 Æ©f =49

By inspection, we can spot that (39.5–49.5) has the highest frequency, i.e. 10. So, this is the modal class.

Thus, the modal age is 42 years.

#### Question 19:

Determine the value of mode for the following data by using the formula:
Mode = 3 Median  − 2   Mean.

 Marks No. of students Less than 10 5 Less than 20 15 Less than 30 98 Less than 40 242 Less than 50 367 Less than 60 405 Less than 70 425 Less than 80 438 Less than 90 439

 Marks c.f. f m fm 0 – 10 5 5 5 25 10 – 20 15 10 15 150 20 – 30 98 83 25 2075 30 – 40 242 144 35 5040 40 – 50 367 125 45 5625 50 – 60 405 38 55 2090 60 – 70 425 20 65 1300 70 – 80 438 13 75 975 80 – 90 439 1 85 85 Æ©f  = 439 Æ©fm =17365

$\overline{X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{17365}{439}=39.55\phantom{\rule{0ex}{0ex}}$
Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{439}{2}\right)}^{\mathrm{th}}$ item, which is the 219.5th item. This corresponds to the class interval 30−40, so this is the median class.

Now,
Mode = 3 Median – 2 Mean
= 3 × 38.43 – 2× 39.55
= 115.29 – 79.1
$\therefore$ Mode = 36.19

#### Question 20:

For the data given below find graphically the following :
(a) First and Third quartile
(b) The central 50% limit of the age .
(c) The number of workers falling in the age group of 28 to 57 years.

 Age in years : 20-24 25-29 30-34 35-39 40-44 45-49 No.of workers : 5 10 15 25 65 40

 Age in years : 50-54 55-59 60-64 65-69 No.of workers : 23 10 5 2

(a)
In order to find quartiles graphically, we first convert the inclusive series into exclusive series and then convert the frequency distribution into cumulative frequency distribution as follows:

 Age Age Frequency (f) Cumulative Frequency (c.f.) 19.5 – 24.5 24.5 – 29.5 29.5 – 34.5 34.5 – 39.5 39.5 – 44.5 44.5 – 49.5 49.5 – 54.5 54.5 – 59.5 59.5 – 64.5 64.5 – 69.5 Less than 24.5 Less than 29.5 Less than 34.5 Less than 39.5 Less than 44.5 Less than 49.5 Less than 54.5 Less than 59.5 Less than 64.5 Less than 69.5 5 10 15 25 65 40 23 10 5 2 5 15 30 55 120 160 183 193 198 200

Thus, the first and third quartiles are 38.5 and 48.25 respectively.

(b) The two limits (i.e. Q1 and Q3) with which central 50% of items lie are 38.5 and 48.25.

(c)

From the graph it is clear that total Number of workers of age 57 years = 190
Total Number of workers of age 28 years = 11
Number of workers falling in the age group of 28 to 57 years = 190 $-$11=179

#### Question 21:

Draw a 'less than' ogive from the following data and hence find out the value of Median.

 Class 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 Frequency 6 9 13 23 19 15 9 6

For constructing a less than ogive, first the given frequency distribution is converted into a less than cumulative frequency distribution as follows:

 Class Frequency (f) Cumulative Frequency (c.f.) Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50 Less than 55 Less than 60 6 9 13 23 19 15 9 6 6 15 28 51 70 85 94 100

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

Hence, median is 39.78.

#### Question 22:

The following table gives the distribution of the wages of 65 employees in a factory.

 Wages (in â‚¹ ) : 50 60 70 80 90 100 110 120 (Equal to more than) Number of employees : 65 57 47 31 17 7 2 0
Draw a 'less than' curve from the above data and estimate the number of employees earning atleast â‚¹ 63 but less than â‚¹ 75.

 Class Interval More than Cumulative Frequency Frequency Wages Less than Cumulative Frequency 50$-$60 60$-$70 70$-$80 80$-$90 90$-$100 100$-$110 110$-$120 120$-$130 65 57 47 31 17 7 2 0 65$-$57= 8 57$-$47=10 47$-$31=16 31$-$17=14 17$-$7= 10 7$-$2=5 2$-$0=2 0 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 Less than 110 Less than 120 Less than 130 8 18 34 48 58 63 65 65

Number of employees earning Rs 75 = 24
Number of employees earning atleast Rs 63 = 10
Number of employees earning between Rs 63 and Rs 75 = 24 $-$10 =14

Therefore, 14 employees are earning between Rs 63 and Rs 75.

#### Question 23:

Draw the histogram and estimate the value of mode from the following data:
â€‹

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No of students 0 2 3 7 13 11 9 2 1

Hence, mode is 47.5.

#### Question 24:

Represent the following data by means of a histogram  and find out mode.

 Weekly wages : 10-15 15-20 20-25 25-30 30-35 35-40 40-45 No of Workers : 7 19 27 15 12 12 8

Hence, mode is 22.

#### Question 25:

For a particular set of data in individual incomes the 'less than type' Ogive and the 'more than type' Ogive were found to intersect at â‚¹ 715. Discuss the significance of 'â‚¹ 715' for the given set of data.

In a graphical presentation of a data, the place where the 'less than type' ogive and 'more than type' ogive intersect is the middle most point of the data. Thus, we may say that the point at which the two ogives intersect represents the median of the given data.

In the given question, the 'less than' and the 'more than' ogives intersect at Rs715.
Therefore, 715 is the median of the given data. It divides the whole data into two  equal parts.

#### Question 26:

Calculate missing frequencies, if median value is 28 and N =44

 Class 0-10 10-20 20-30 30-40 40-50 50-60 No of students 3 ? 15 9 ? 4

 Class Frequency (f) m c.f 0 – 10 3 5 3 10 – â€‹20 ? (f1) 15 3 + f1 20 – 30 15 25 18 + f1 30 – 40 9 35 27 + f1 40 – 50 ? (f2) 45 27 + f1 + f2 50 – 60 4 55 44 44

We know, total frequency = 44
So, 31 + f1 f2 = 44
or, f1 + f2 = 13

As, we know f1 + f2 = 13, and f1 is 7, so, f2 becomes (13-7 =) 6 .

#### Question 27:

Calculate Median from the data given below:

 Class Interval 0-10 10-20 20-40 40-60 60-80 80-100 Frequency 10 8 22 10 20 10

 Class Frequency (f) c.f 0 – 10 10 10 10 – â€‹20 8 18 20 – 40 22 40 40 – 60 10 50 60 – 80 20 70 80 – 100 10 80 80

#### Question 28:

Calculate mode from  the following data:

 Marks (more than) 0-9 10-19 20-29 30-39 40-49 50-59 No of students 2 5 15 25 10 3

 X f 0 – 9.5 2 9.5 – 19.5 5 19.5 – 29.5 15 29.5 – 39.5 25 $\underset{}{\to }$Modal class 39.5 – 49.5 10 49.5 – 59.5 3

By inspection, we can spot that (29.5–39.5) has the highest frequency, i.e. 25. So, this is the modal class.

Thus, the mode is 33.5.

#### Question 29:

Calculate mode from the following data:

 Class Interval 0-10 10-20 20-30 30-40 40-50 50-70 Frequency 7 10 8 20 32 28

â€‹

 X f 0 – 10 7 10 – 20 10 20 – 30 8 30 – 40 20 40 – 50 32 $\underset{}{\to }$Modal class 50 – 70 28

By inspection, we can spot that (40 – 50) has the highest frequency, i.e. 32. So, this is the modal class.

Thus, the mode is 44.
Note, f2 will be 14 as the adjusted frequency for the class 50 - 70 is 14.

#### Question 30:

Calculate mode from following data:

 Class Interval 5-14 15-24 25-34 35-44 44-54 55-64 Frequency 8 10 15 25 40 20

 X f 4.5 – 14.5 8 14.5 – 24.5 10 24.5 – 34.5 15 34.5 – 44.5 25 44.5 – 54.5 40 $\underset{}{\to }$Modal class 54.5 – 64.5 20