GSEB Class 11 Statistics Solutions Chapter 3 Measures of Central Tendency Exercise 3.1

GSEB Solutions Class 11 Statistics Chapter 3 Measures of Central Tendency Ex 3.1

 

Question 1. The weekly growths (in cm) of saplings in a nursery are: 1.0, 3.2, 1.4, 1.9, 2.4, 1.6, 1.4, 2.1, 1.3, 1.5. Find the mean growth.
Answer:Here, the total number of observations, \(n = 10\), represents the weekly growth (in cm) of plants. The mean growth of the plant is calculated as follows: \[ \bar{x} = \frac{\Sigma x}{n} \] \[ = \frac{1.0+3.2+1.4+1.9+2.4+1.6+1.4+2.1+1.3+1.5}{10} \] \[ = \frac{17.8}{10} \] \[ = 1.78 \text{ cm} \]In simple words: The mean growth is found by summing all individual growth values and then dividing by the total count of saplings, which results in an average growth of 1.78 cm.

🎯 Exam Tip: When calculating the mean for ungrouped data, ensure all data points are accurately summed before dividing by the total number of observations. Double-check your addition.

 

Question 2. The mean age of 4 participants in a relay race was calculated to be 24 years. Later it was found that one of the participant's age was actually 27 years, which was wrongly recorded as 25 years. If there is a rule wherein it is not possible to participate in this race if the mean age is more than 25 years, can they participate in this race even after the correction of age?
Answer:Initially, the mean age of 4 participants was 24 years.
\( \implies \) The sum of ages for the 4 participants = \(4 \times 24 = 96\) years. It was discovered that one participant's age was 27 years, but incorrectly recorded as 25 years. To correct this, the true sum of ages for the 4 participants = \(96 - 25 + 27 = 98\) years. The corrected mean age for the 4 participants = \( \frac{98}{4} = 24.5 \) years. According to the rule, participants cannot join if the mean age exceeds 25 years. Since the corrected mean age is 24.5 years, which is less than 25 years, the participants are eligible to participate in the race.In simple words: After correcting a data entry error, the team's average age dropped from 24 years to 24.5 years, which is still below the 25-year participation limit, so they can compete.

🎯 Exam Tip: In problems involving corrected means, always first calculate the 'true sum' by subtracting the incorrect value and adding the correct value. Then, recalculate the mean using this true sum to determine eligibility or final status.

 

Question 3. The following table gives the diameters (in mm) of different screws selected from a large consignment. Find the mean diameter.

Diameter of screw (mm)	30	35	40	45	50	55
No. of screws	4	10	15	8	5	3

Answer:Here, \(x\) represents the diameter (in mm) of the screw, and \(f\) denotes the number of screws. The total number of screws selected, \(n\), is equivalent to the sum of frequencies (\( \Sigma f \)). The calculation for the mean diameter is detailed in the table below:

Diameter of the screw (mm) x	No. of screws f	f.x
30	4	120
35	10	350
40	15	600
45	8	360
50	5	250
55	3	165
Total	n = 45	\( \Sigma fx = 1845 \)

The mean of the diameter of the screws is: \[ \bar{x} = \frac{\Sigma fx}{n} \] \[ = \frac{1845}{45} \] \[ = 41 \text{ mm} \]In simple words: By multiplying each screw diameter by its frequency, summing these products, and dividing by the total number of screws, the average diameter is found to be 41 mm.

🎯 Exam Tip: When calculating the mean for grouped discrete data, accurately compute the product of each 'x' and 'f' value, then sum these products to get \( \Sigma fx \). This is crucial for avoiding calculation errors.

 

Question 4. The marks in a test for a group of students are as follows. Find the mean marks of these students.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70
No. of students	3	5	12	16	11	5	4

Answer:The provided frequency distribution features classes of equal length. Therefore, the mean can be efficiently calculated using the short-cut method, where \( d = \frac{x-A}{c} \). The complete calculation table is presented below:

Marks	No. of students f	Mid value x	\( d = \frac{x-A}{c} \) A=35, c=10	f.d
0-10	3	5	-3	-9
10-20	5	15	-2	-10
20-30	12	25	-1	-12
30-40	16	35	0	0
40-50	11	45	1	11
50-60	5	55	2	10
60-70	4	65	3	12
Total	n = 56	-	-	\( \Sigma fd = 2 \)

The mean of students' marks is calculated using the formula: \[ \bar{x} = A + \frac{\Sigma fd}{n} \times c \] Substituting the values \(A = 35\), \( \Sigma fd = 2\), \(n = 56\), and \(c = 10\) into the formula: \[ \bar{x} = 35 + \frac{2}{56} \times 10 \] \[ \bar{x} = 35 + \frac{20}{56} \] \[ \bar{x} = 35 + 0.3571 \] \[ \bar{x} \approx 35.36 \text{ marks} \]In simple words: Using the shortcut method for grouped data, where the assumed mean is 35 and class interval is 10, the calculated average mark for the students is approximately 35.36.

🎯 Exam Tip: When using the shortcut method for calculating mean, accurately determine the mid-point of each class, choose an appropriate assumed mean (A), and correctly calculate the deviation (d) and \( \Sigma fd \). Mistakes in these initial steps will lead to an incorrect final mean.

 

Question 5. The following information is available on the talk time (in min.) noted for 70 calls of a certain mobile phone user. Find the mean talk time.

Talk time (in min.)	Less than 4	Less than 8	Less than 12	Less than 16	Less than 20
No. of calls	20	42	57	65	70

Answer:The given frequency distribution is a 'less than' type cumulative frequency distribution. The difference between successive upper boundary points is 4, which indicates a class length \(c = 4\). This also implies the lower boundary point of the initial class is 0. The frequency of each class is derived from the given cumulative frequency data. The reconstructed frequency distribution and the mean calculation are shown in the table below:

Talk time (minute)	No. of calls f	Mid value x	\( d = \frac{x-A}{c} \) A=10, c=4	f.d
0-4	20	2	-2	-40
4-8	22	6	-1	-22
8-12	15	10	0	0
12-16	8	14	1	8
16-20	5	18	2	10
Total	n = 70	-	-	\( \Sigma fd = -44 \)

The mean of the talk time is calculated as follows: \[ \bar{x} = A + \frac{\Sigma fd}{n} \times c \] Substituting \(A = 10\), \( \Sigma fd = -44\), \(n = 70\), and \(c = 4\) into the formula: \[ \bar{x} = 10 + \frac{-44}{70} \times 4 \] \[ \bar{x} = 10 - \frac{176}{70} \] \[ \bar{x} = 10 - 2.514 \] \[ \bar{x} \approx 7.49 \text{ minutes} \] Thus, the mean talk time is approximately 7.49 minutes.In simple words: By converting the cumulative frequency data into a regular frequency distribution and applying the short-cut method, the average call duration for the mobile phone user is found to be about 7.49 minutes.

🎯 Exam Tip: When dealing with 'less than' or 'more than' cumulative frequency distributions, the first step is to convert them into standard frequency distributions with clear class intervals. Ensure accurate calculation of frequencies for each class before proceeding with mean calculation.

 

Question 6. The distribution of profits (in lakh Rs.) of 50 firms is given below. Find the mean profit.

Profit (lakh Rs.)	0-7	7-14	14-21	21-28	28-35
No. of firms	4	9	18	12	7

Answer:The provided frequency distribution has equal class lengths. Therefore, we will calculate the mean using the short-cut method, with \( d = \frac{x-A}{c} \). The calculation table is prepared as follows:

Profit (lakh Rs.)	No. of firms f	Mid value x	\( d = \frac{x-A}{c} \) A=17.5, c=7	f.d
0-7	4	3.5	-2	-8
7-14	9	10.5	-1	-9
14-21	18	17.5	0	0
21-28	12	22.5	1	12
28-35	7	27.5	2	14
Total	n = 50	-	-	\( \Sigma fd = 9 \)

The mean profit is calculated using the formula: \[ \bar{x} = A + \frac{\Sigma fd}{n} \times c \] Substituting \(A = 17.5\), \( \Sigma fd = 9\), \(n = 50\), and \(c = 7\) into the formula: \[ \bar{x} = 17.5 + \frac{9}{50} \times 7 \] \[ \bar{x} = 17.5 + \frac{63}{50} \] \[ \bar{x} = 17.5 + 1.26 \] \[ \bar{x} = 18.76 \text{ lakh Rs.} \] Hence, the mean profit is 18.76 lakh Rs.In simple words: Applying the short-cut method to the distribution of profits, with an assumed mean of 17.5 and a class width of 7, reveals that the average profit for these firms is 18.76 lakh Rs.

🎯 Exam Tip: For continuous frequency distributions with equal class intervals, using the short-cut method (step-deviation method) simplifies calculations significantly. Ensure the assumed mean (A) and class length (c) are correctly identified to prevent errors.

 

Question 7. The distribution of demand of an item on different days is as follows. Find the mean demand.

Demand (units)	5-14	15-24	25-34	35-49	50-64	65-79
No. of days	4	17	19	22	18	10

Answer:The given inclusive continuous frequency distribution has unequal class lengths. The first three classes have a length of 10, while the subsequent three classes have a length of 15. Since 5 is a common factor for both 10 and 15, we will use \( d = \frac{x-A}{5} \) to calculate the mean via the short-cut method. The calculation table is prepared as follows:

Demand (units)	No. of days f	Mid value x	\( d = \frac{x-A}{c} \) A=29.5, c=5	f.d
5-14	4	9.5	-4	-16
15-24	17	19.5	-2	-34
25-34	19	29.5	0	0
35-49	22	42.0	2.5	55
50-64	18	57.0	5.5	99
65-79	10	72.0	8.5	85
Total	n = 90	-	-	\( \Sigma fd = 189 \)

The mean of the demand is calculated using the formula: \[ \bar{x} = A + \frac{\Sigma fd}{n} \times c \] Substituting \(A = 29.5\), \( \Sigma fd = 189\), \(n = 90\), and \(c = 5\) into the formula: \[ \bar{x} = 29.5 + \frac{189}{90} \times 5 \] \[ \bar{x} = 29.5 + \frac{945}{90} \] \[ \bar{x} = 29.5 + 10.5 \] \[ \bar{x} = 40 \text{ units} \] Therefore, the mean demand is 40 units.In simple words: Despite unequal class intervals, by selecting a common factor for class length and applying the short-cut method, the average demand for the item across different days is found to be 40 units.

🎯 Exam Tip: For frequency distributions with unequal class lengths but a common factor for class intervals, it's crucial to correctly identify this common factor to serve as 'c' in the step-deviation method. Ensure all mid-values and deviations are calculated precisely.