Mean & Median of Ungrouped Data & Frequency Polygon — Competency Focused Questions

Which of the following statements is/are correct?

(i) Mean of a data can be smaller than the smallest number of the data.

(ii) Mean of a data can be bigger than the largest number of the data.

1. Only (i)

2. Only (ii)

3. Both (i) and (ii)

4. Both are wrong

Answer

(i) If we have a set of numbers, the average cannot be lower than the lowest value because there is no data pulling the average down past that point.

∴ Statement (i) is wrong.

(ii) Mean of a data cannot be bigger than the larger number because the average represents a "middle" value. So, their average cannot mathematically go beyond that value.

∴ Statement (ii) is wrong.

Both are wrong.

Hence, option 4 is the correct option.

Average marks obtained by students of a class in an exam is 70. Which of the following is definitely incorrect?

(i) One of the students scored more than 100 marks.

(ii) One of the students scored less than 10 marks.

1. Only (i)

2. Only (ii)

3. Both (i) and (ii)

4. None of these

Answer

Average Mean marks = 70

Total marks = 70 × Number of students

(i) This can be possible.

E.g : Suppose 3 students scored :

120, 50, 40

Mean = $\dfrac{210}{3}$ = 70.

So, scoring more than 100 is possible.

∴ Statement (i) is wrong.

(ii) Scoring less than 10 marks is also possible.

E.g : Suppose a student scored :

0, 110, 100

Mean = $\dfrac{210}{3}$ = 70.

So, scoring less than 10 marks is also possible.

∴ Statement (ii) is wrong.

Both are wrong.

Hence, option 4 is the correct option.

A government agency collected the data of individual incomes of 10,625 people and decided that the individual that have less income than the median of the data are economically poor. Assuming that all the individuals have different incomes, which of the following is correct?

1. The number of economically poor people is less than the number of rich people.

2. The number of economically poor people is more than the number of rich people.

3. Exactly half of the people are economically poor.

4. We can't say anything about the number of economically poor people.

Answer

Total people = 10,625

Since, number of observations is odd

∴ Median = $\dfrac{n + 1}{2}$

= $\dfrac{10625 + 1}{2}$

= $\dfrac{10626}{2}$ = 5313.

So median is the 5313th person.

Number of economically poor people are = 5313 - 1 = 5312.

This means that number of rich people are = 10625 - 5312 = 5313.

∴ Number of economically poor people is less than the number of rich people.

Hence, option 1 is the correct option.

If the mean of a and $\dfrac{1}{a}$ is x, then the mean of a³ and $\dfrac{1}{a^3}$ is :

1. x(4x² - 3)

2. x(x²- 2)

3. x² + 3

4. x²

Answer

Mean of a and $\dfrac{1}{a}$ is x

$\Rightarrow \dfrac{a + \dfrac{1}{a}}{2} = x$

$\Rightarrow a + \dfrac{1}{a} = 2x$

We know that,

(a + b)³ = a³ + b³ + 3ab(a + b)

Substituting $b = \dfrac{1}{a}$ :

$\Rightarrow \left(a + \dfrac{1}{a}\right)^3 = a^3 + \dfrac{1}{a^3} + 3(a)\left(\dfrac{1}{a}\right)\left(a + \dfrac{1}{a}\right) \\[1em] \Rightarrow \left(a + \dfrac{1}{a}\right)^3 = a^3 + \dfrac{1}{a^3} + 3\left(a + \dfrac{1}{a}\right) \\[1em]$

Substituting $a + \dfrac{1}{a} = 2x$ into the identity:

$\Rightarrow (2x)^3 = a^3 + \dfrac{1}{a^3} + 3(2x) \\[1em] \Rightarrow 8x^3 = a^3 + \dfrac{1}{a^3} + 6x \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 8x^3 - 6x \\[1em]$

The mean of a³ and $\dfrac{1}{a^3}$ is:

$\Rightarrow \text{Mean} = \dfrac{a^3 + \dfrac{1}{a^3}}{2} \\[1em] \Rightarrow \text{Mean} = \dfrac{8x^3 - 6x}{2} \\[1em] \Rightarrow \text{Mean} = 4x^3 - 3x \\[1em] \Rightarrow \text{Mean} = x(4x^2 - 3)$

Hence, option 1 is the correct option.

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is :

1. 49.5

2. 53

3. 46.5

4. 56.5

Answer

Let the original numbers be :

x₁, x₂, x₃, ....., x₅₀

And their mean be 'a'

So,

$\Rightarrow a = \dfrac{x_1 + x_2 + \dots + x_{50}}{50} \\[1em] \Rightarrow x_1 + x_2 + \dots + x_{50} = 50a \\[1em]$

Each number is subtracted from 53.

∴ The new numbers are :

$(53 - x_1), (53 - x_2), (53 - x_3), \dots, (53 - x_{50})$

Sum of new numbers is :

⇒ $(53 - x_1) + (53 - x_2) + (53 - x_3) + \dots + (53 - x_{50})$

⇒ 53 + 53 + ..... + 53 - (x₁ + x₂ + x₃ + ..... + x₅₀)

⇒ 50 × 53 - 50a

⇒ 2650 - 50a

Mean = $\dfrac{\text{Total Sum}}{\text{Total observations}}$

⇒ -3.5 = $\dfrac{2650 - 50a}{50}$

⇒ -175 = 2650 - 50a

⇒ 50a = 2650 + 175

⇒ a = $\dfrac{2825}{50}$

⇒ a = 56.5

∴ Mean of original numbers = 56.5.

Hence, option 4 is the correct option.

There are 10 observations x₁, x₂, x₃, ....., x₁₀ in a data. The first five elements x₁, x₂, ....., x₅ are placed by x₁ + 5, x₂ + 5, ....., x₅ + 5 respectively. Also, the next five elements x₆, x₇, ....., x₁₀ are placed by x₆ - 5, x₇ - 5, ..., x₁₀ - 5 respectively. Calculate the overall change in the mean of the data.

Answer

Total observations = 10

First 5 numbers increased by 5

∴ Total increase in sum = 5 × 5 = 25

Next 5 numbers decreased by 5

∴ Total decrease in sum = 5 × 5 = 25

So, net change = 25 - 25 = 0

So new total sum = original total sum.

Hence, there is no change in the mean of a data.

In a collection of data, there are 5 numbers and the numbers in the data are 3, 4, 12 and 13. If the median is 12, then what can be maximum value of mean of the data?

Answer

Total numbers = 5

For 5 numbers :

Median = $\dfrac{n + 1}{2}$ observation

⇒ 3rd observation.

For median to be 12

The arrangement should be :

3, 4, 12, 13, x or 3, 4, 12, x, 13.

In order to maximum mean, the value of x should be greatest, thus arrangement will be :

3, 4, 12, 13, x

Since, one of the values must is repeated, so value of x will be equal to 13.

3, 4, 12, 13, 13

Mean = $\dfrac{3 + 4 + 12 + 13 + 13}{5} = \dfrac{45}{5}$ = 9.

Hence, maximum mean value = 9.

In an office, mean of salaries of employees is ₹1,00,000 and median is ₹80,000. An employee will be considered, Grade A officer if his income is at least ₹85,000. Based on the above information check the validity of the following statements.

(i) Number of Grade A officers is more than or equal to number of non Grade A officers.

(ii) Mean of salaries of Grade A officers is more than ₹1,00,000.

Answer

Mean = ₹1,00,000

Median (mid value) = ₹80,000

This means:

50% employees earn ≤ ₹80,000

50% employees earn ≥ ₹80,000

(i) To be a Grade A officer, an employee must earn at least ₹85,000.

Since the income (₹85,000) is higher than the median (₹80,000), So fewer than 50% of the employees can qualify for Grade A.

∴ There will be more non Grade A officers than the Grade A officers.

Hence, statement (i) is false.

(ii) Given data shows that Mean (₹1,00,000) is significantly higher than the Median (₹80,000). This means there is a small group of very high earners pulling the average up.

The overall mean (₹1,00,000) includes non Grade A officers who earn less than ₹85,000. Therefore there group mean will definitely be below ₹85,000.

To pull this anchor of the non-Grade A officers up from less than ₹85,000 to a total average of ₹1,00,000, the Grade A officers must have a mean significantly higher than ₹1,00,000.

∴ Mean salaries of Grade A officers is more than ₹1,00,000.

Hence, statement (ii) is true.