Statistics

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean and why ?

Answer

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

We will use direct method for solving the mean of the above data as the numerical values of x_i and f_i are small.

By formula,

Mean = $\dfrac{Σf_ix_i}{Σf_i} = \dfrac{162}{20}$ = 8.1

Hence, mean number of plants in each house = 8.1

Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = $\dfrac{Σf_ix_i}{Σf_i} = \dfrac{27260}{50}$ = 545.20

Hence, mean daily wage = ₹ 545.20

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Answer

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = $\dfrac{Σf_ix_i}{Σf_i}$

Substituting values we get :

$\Rightarrow 18 = \dfrac{752 + 20f}{44 + f} \\[1em] \Rightarrow 18(44 + f) = 752 + 20f \\[1em] \Rightarrow 792 + 18f = 752 + 20f \\[1em] \Rightarrow 20f - 18f = 792 - 752 \\[1em] \Rightarrow 2f = 40 \\[1em] \Rightarrow f = \dfrac{40}{2} = 20.$

Hence, f = 20.

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Answer

We will use step deviation method.

In the following table a is the assumed mean and h is the class size.

Here, h = 3.

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = a + $\dfrac{Σf_iu_i}{Σf_i} \times h$

Substituting values we get :

$\text{Mean } = 75.5 + \dfrac{4}{30} \times 3 \\[1em] = 75.5 + \dfrac{4}{10} \\[1em] = 75.5 + 0.4 \\[1em] = 75.9$

Hence, mean = 75.9

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?

Answer

Here the given data is discontinuous.

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\dfrac{53 - 52}{2} = \dfrac{1}{2}$ = 0.5

∴ 0.5 has to be added to the upper-class limit and 0.5 has to be subtracted from the lower-class limit of each interval.

We will use step deviation method to find the mean.

In the following table a is the assumed mean and h is the class size.

Here, h = 3.

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = a + $\dfrac{Σf_iu_i}{Σf_i} \times h$

Substituting values we get :

$\text{Mean } = 57 + \dfrac{25}{400} \times 3 \\[1em] = 57 + \dfrac{3}{16} \\[1em] = 57 + 0.19 \\[1em] = 57.19$

Hence, mean number of mangoes = 57.19

The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Answer

We will use step deviation method to find the mean.

In the following table a is the assumed mean and h is the class size.

Here, h = 50.

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = a + $\dfrac{Σf_iu_i}{Σf_i} \times h$

Substituting values we get :

$\text{Mean } = 225 + \dfrac{-7}{25} \times 50 \\[1em] = 225 + (-14) \\[1em] = 211$

Hence, mean daily expenditure on food = ₹ 211.

To find out the concentration of SO₂ (in parts per million, i.e., ppm), in the air, the data was collected for 30 localities in a certain city and is presented below :

Find the mean concentration of SO₂ in the air.

Answer

We will use step deviation method to find the mean.

In the following table a is the assumed mean and h is the class size.

Here, h = 0.04

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = a + $\dfrac{Σf_iu_i}{Σf_i} \times h$

Substituting values we get :

$\text{Mean } = 0.10 + \dfrac{-1}{30} \times 0.04 \\[1em] = 0.10 - \dfrac{0.04}{30} \\[1em] = 0.10 - \dfrac{4}{3000} \\[1em] = 0.10 - \dfrac{1}{750} \\[1em] = 0.10 - 0.00133 \\[1em] = 0.099$

Hence, mean concentration of SO₂ = 0.099 ppm.

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Answer

We will use assumed mean method to find the mean. Here, a is the assumed mean.

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = $a + \dfrac{Σf_id_i}{Σf_i}$

Substituting values we get :

$\text{Mean } = 17 + \dfrac{-181}{40} \\[1em] = 17 - \dfrac{181}{40} \\[1em] = 17 - 4.525 \\[1em] = 12.475 ≈ 12.48$

Hence, mean number of days = 12.48

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer

We will use assumed mean method to find the mean. Here, a is the assumed mean.

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = $a + \dfrac{Σf_id_i}{Σf_i}$

Substituting values we get :

$\text{Mean } = 70 + \dfrac{-20}{35} \\[1em] = 70 - \dfrac{20}{35} \\[1em] = 70 - 0.57 \\[1em] = 69.43$

Hence, mean literacy rate = 69.43 %.

The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer

We will find mean by direct method.

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = $\dfrac{Σf_ix_i}{Σf_i} = \dfrac{2830}{80} = 35.375$

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

Class 35 - 45 has the highest frequency.

∴ It is the modal class.

∴ l = 35, f₁ = 23, f₀ = 21, f₂ = 14 and h = 10.

Substituting values we get :

$\text{Mode} = 35 + \Big(\dfrac{23 - 21}{2 \times 23 - 21 - 14}\Big) \times 10 \\[1em] = 35 + \dfrac{2}{46 - 35} \times 10 \\[1em] = 35 + \dfrac{20}{11} \\[1em] = 35 + 1.8 \\[1em] = 36.8$

Since, mode = 36.8

∴ Maximum number of patients admitted in the hospital are of the age 36.8 years.

Since, mean = 35.37

∴ Average the age of a patient admitted to the hospital is 35.37 years.

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Answer

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

Class 60 - 80 has the highest frequency.

∴ It is the modal class.

∴ l = 60, f₁ = 61, f₀ = 52, f₂ = 38 and h = 20.

Substituting values we get :

$\text{Mode} = 60 + \Big(\dfrac{61 - 52}{2 \times 61 - 52 - 38}\Big) \times 20 \\[1em] = 60 + \dfrac{9}{122 - 90} \times 20 \\[1em] = 60 + \dfrac{180}{32} \\[1em] = 60 + 5.625 \\[1em] = 65.625$

Hence, modal lifetime = 65.625 hours.

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Answer

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

Class 1500 - 2000 has the highest frequency.

∴ It is the modal class.

∴ l = 1500, f₁ = 40, f₀ = 24, f₂ = 33 and h = 500.

Substituting values we get :

$\text{Mode} = 1500 + \Big(\dfrac{40 - 24}{2 \times 40 - 24 - 33}\Big) \times 500 \\[1em] = 1500 + \dfrac{16}{80 - 57} \times 500 \\[1em] = 1500 + \dfrac{16}{23} \times 500 \\[1em] = 1500 + \dfrac{8000}{23} \\[1em] = 1500 + 347.83 \\[1em] = 1847.83$

We will find mean using step deviation method.

By formula,

Mean = a + $\dfrac{Σf_iu_i}{Σf_i} \times h$

Substituting values we get :

$\text{Mean } = 3250 + \dfrac{-235}{200} \times 500 \\[1em] = 3250 - \dfrac{1175}{2} \\[1em] = 3250 - 587.5 \\[1em] = 2662.50$

Hence, mean = ₹ 2662.50 and mode = ₹ 1847.83.

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Answer

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

Class 30 - 35 has the highest frequency.

∴ It is the modal class.

∴ l = 30, f₁ = 10, f₀ = 9, f₂ = 3 and h = 5.

Substituting values we get :

$\text{Mode} = 30 + \Big(\dfrac{10 - 9}{2 \times 10 - 9 - 3}\Big) \times 5 \\[1em] = 30 + \dfrac{1}{20 - 12} \times 5 \\[1em] = 30 + \dfrac{5}{8} \\[1em] = 30 + 0.625 \\[1em] = 30.625 ≈ 30.6$

We will find mean using direct method.

By formula,

Mean = $\dfrac{Σf_ix_i}{Σf_i} = \dfrac{1022.5}{35}$ = 29.2

Hence, mode = 30.6 and median = 29.2 so, most states/U.T. have a student teacher ratio of 30.6 and on an average, this ratio is 29.2

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Answer

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

Class 4000 - 5000 has the highest frequency.

∴ It is the modal class.

∴ l = 4000, f₁ = 18, f₀ = 4, f₂ = 9 and h = 1000.

Substituting values we get :

$\text{Mode} = 4000 + \Big(\dfrac{18 - 4}{2 \times 18 - 4 - 9}\Big) \times 1000 \\[1em] = 4000 + \dfrac{14}{36 - 13} \times 1000 \\[1em] = 4000 + \dfrac{14000}{23} \\[1em] = 4000 + 608.7 \\[1em] = 4608.7$

Hence, mode = 4608.7 runs.

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Answer

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

Class 40 - 50 has the highest frequency.

∴ It is the modal class.

∴ l = 40, f₁ = 20, f₀ = 12, f₂ = 11 and h = 10.

Substituting values we get :

$\text{Mode} = 40 + \Big(\dfrac{20 - 12}{2 \times 20 - 12 - 11}\Big) \times 10 \\[1em] = 40 + \dfrac{8}{40 - 23} \times 10 \\[1em] = 40 + \dfrac{80}{17} \\[1em] = 40 + 4.7 \\[1em] = 44.7$

Hence, mode = 44.7 cars.

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Answer

We will find mean by step deviation method.

By formula,

Class mark = $\dfrac{\text{Lower limit + Upper limit}}{2}$

Here, h (class size) = 20.

By formula,

Mean = $a + \dfrac{Σf_iu_i}{Σf_i} \times h$

Substituting values we get :

$\text{Mean } = 135 + \dfrac{7}{68} \times 20 \\[1em] = 135 + \dfrac{35}{17} \\[1em] = 135 + 2.05 = 137.05$

Cumulative frequency distribution table is as follows :

Here, n = 68, which is even

$\dfrac{n}{2} = \dfrac{68}{2}$ = 34.

Cumulative frequency just greater than $\dfrac{n}{2}$ is 42, belonging to class-interval 125 - 145.

∴ Median class = 125 - 145

⇒ Lower limit of median class (l) = 125

⇒ Frequency of median class (f) = 20

⇒ Cumulative frequency of class preceding median class (cf) = 22

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\text{Median} = 125 + \dfrac{34 - 22}{20} \times 20 \\[1em] = 125 + \dfrac{12}{20} \times 20 \\[1em] = 125 + 12 \\[1em] = 137.$

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

From table,

Class 125 - 145 has the highest frequency.

∴ It is the modal class.

∴ l = 125, f₁ = 20, f₀ = 13, f₂ = 14 and h = 20.

Substituting values we get :

$\text{Mode} = 125 + \Big(\dfrac{20 - 13}{2 \times 20 - 13 - 14}\Big) \times 20 \\[1em] = 125 + \dfrac{7}{40 - 27} \times 20 \\[1em] = 125 + \dfrac{140}{13} \\[1em] = 125 + 10.76 \\[1em] = 135.76$

Hence, mean = 137.05, median = 137 and mode = 135.76

If the median of the distribution given below is 28.5, find the values of x and y.

Answer

Cumulative frequency distribution table is as follows :

We know that,

⇒ n = 60

⇒ 45 + x + y = 60

⇒ x + y = 15 ...........(1)

Given,

Median = 28.5

From cumulative frequency distribution table we get :

Median lies in class 20 - 30.

∴ Median class = 20 - 30

⇒ Lower limit of median class (l) = 20

⇒ Class size (h) = 10

⇒ Frequency of median class (f) = 20

⇒ Cumulative frequency of class preceding median class (cf) = 5 + x

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\Rightarrow 28.5 = 20 + \Big(\dfrac{\dfrac{60}{2} - (5 + x)}{20}\Big) \times 10 \\[1em] \Rightarrow 28.5 - 20 = \dfrac{30 - 5 - x}{2} \\[1em] \Rightarrow 8.5 \times 2 = 25 - x \\[1em] \Rightarrow 17 = 25 - x \\[1em] \Rightarrow x = 25 - 17 = 8.$

Substituting value of x in equation (1), we get :

⇒ 8 + y = 15

⇒ y = 15 - 8

⇒ y = 7.

Hence, x = 8 and y = 7.

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Answer

Cumulative frequency distribution table is as follows :

Here, n = 100, $\dfrac{n}{2} = 50$.

Cumulative frequency just greater than 50 is 78, belonging to class-interval 35 − 40.

Therefore, median class = 35 - 40

⇒ Class size (h) = 5

⇒ Lower limit of median class (l) = 35

⇒ Frequency of median class (f) = 33

⇒ Cumulative frequency of class preceding median class (cf) = 45

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\Rightarrow \text{Median } = 35 + \dfrac{50 - 45}{33} \times 5 \\[1em] = 35 + \dfrac{25}{33} \\[1em] = 35 + 0.76 \\[1em] = 35.76$

Hence, median age = 35.76 years.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Find the median length of leaves.

Answer

Here the given data is discontinuous.

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\dfrac{127 - 126}{2} = \dfrac{1}{2}$ = 0.5

∴ 0.5 has to be added to the upper-class limit and 0.5 has to be subtracted from the lower-class limit of each interval.

Cumulative frequency distribution table is as follows :

Here, n = 40, $\dfrac{n}{2} = \dfrac{40}{2} = 20$.

Cumulative frequency just greater than 20 is 29, belonging to class-interval 144.5 − 153.5

Therefore, median class = 144.5 - 153.5

⇒ Class size (h) = 9

⇒ Lower limit of median class (l) = 144.5

⇒ Frequency of median class (f) = 12

⇒ Cumulative frequency of class preceding median class (cf) = 17

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\Rightarrow \text{Median } = 144.5 + \dfrac{20 - 17}{12} \times 9 \\[1em] = 144.5 + \dfrac{27}{12} \\[1em] = 144.5 + 2.25 \\[1em] = 146.75$

Hence, median length = 146.75 mm.

The following table gives the distribution of the life time of 400 neon lamps :

Find the median life time of a lamp.

Answer

Cumulative frequency distribution table is as follows :

Here, n = 400, $\dfrac{n}{2} = \dfrac{400}{2} = 200$.

Cumulative frequency just greater than 200 is 216, belonging to class-interval 3000 - 3500

∴ Median class = 3000 - 3500

⇒ Class size (h) = 500

⇒ Lower limit of median class (l) = 3000

⇒ Frequency of median class (f) = 86

⇒ Cumulative frequency of class preceding median class (cf) = 130

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\Rightarrow \text{Median } = 3000 + \dfrac{200 - 130}{86} \times 500 \\[1em] = 3000 + \dfrac{35000}{86} \\[1em] = 3000 + 406.98 \\[1em] = 3406.98$

Hence, median life = 3406.98 hours.

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Answer

We will use direct method to find the mean.

By formula,

Mean = $\dfrac{Σf_ix_i}{Σf_i} = \dfrac{832}{100}$ = 8.32

Cumulative frequency distribution table is as follows :

Here, n = 100, $\dfrac{n}{2} = \dfrac{100}{2}$ = 50.

Cumulative frequency just greater than 50 is 76, belonging to class-interval 7 - 10.

∴ Median class = 7 - 10

⇒ Class size (h) = 3

⇒ Lower limit of median class (l) = 7

⇒ Frequency of median class (f) = 40

⇒ Cumulative frequency of class preceding median class (cf) = 36

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\text{Median} = 7 + \dfrac{50 - 36}{40} \times 3 \\[1em] = 7 + \dfrac{14 \times 3}{40} \\[1em] = 7 + \dfrac{42}{40} \\[1em] = 7 + 1.05 \\[1em] = 8.05$

By formula,

Mode = l + $\Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h$

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

From table,

Class 7 - 10 has the highest frequency.

∴ It is the modal class.

∴ l = 7, f₁ = 40, f₀ = 30, f₂ = 16 and h = 3.

Substituting values we get :

$\text{Mode} = 7 + \Big(\dfrac{40 - 30}{2 \times 40 - 30 - 16}\Big) \times 3 \\[1em] = 7 + \dfrac{10}{80 - 46} \times 3 \\[1em] = 7 + \dfrac{30}{34} \\[1em] = 7 + 0.88 \\[1em] = 7.88$

Hence, mean = 8.32, median = 8.05 and mode = 7.88

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer

Cumulative frequency distribution table is as follows :

Here, n = 30, $\dfrac{n}{2} = \dfrac{30}{2} = 15$.

Cumulative frequency just greater than 15 is 19, belonging to class-interval 55 - 60

∴ Median class = 55 - 60

Class size (h) = 5

Lower limit of median class (l) = 55

Frequency of median class (f) = 6

Cumulative frequency of class preceding median class (cf) = 13

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\Rightarrow \text{Median } = 55 + \dfrac{15 - 13}{6} \times 5 \\[1em] = 55 + \dfrac{10}{6} \\[1em] = 55 + 1.67 \\[1em] = 56.67$

Hence, median weight = 56.67 kg.