## RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data,Cumulative Frequency Graph and O give MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data,Cumulative Frequency Graph and O give MCQ.

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9a
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9b
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9c
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9d
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9e
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9f
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data MCQ
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself

**Question 1.**

**Solution:
**Correct option: (d)

Range is not a measure of central tendency.

**Question 2.**

**Solution:
**Correct option: (a)

Mean cannot be determined graphically.

**Question 3.**

**Solution:
**Correct option: (a)

Since mean is the average of all observations, it is influenced by extreme values.

**Question 4.**

**Solution:
**Correct option: (c)

Mode can be obtained graphically from a histogram.

**Question 5.**

**Solution:
**Correct option: (d)

Ogives are used to determine the median of a frequency distribution.

**Question 6.**

**Solution:
**Correct option: (b)

The cumulative frequency table is useful in determining the median.

**Question 7.**

**Solution:
**Correct option: (b)

Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.

**Question 8.**

**Solution:
**Correct option: (b)

For a grouped data,

\(\sum { f }_{ i }\left( { x }_{ i }-\bar { x } \right) =0\)

**Question 9.**

**Solution:
**Correct option: (b)

By formula method,

\(\bar { x } =\quad A+h\left[ \frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right] \)

where \({ u }_{ i }=\quad \frac { \left( { x }_{ i }-A \right) }{ h } \)

**Question 10.**

**Solution:
**Correct option: (c)

d

_{i}‘s are the deviations from A of midpoints of the classes.

**Question 11.**

**Solution:
**Correct option: (b)

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.

**Question 12.**

**Solution:
**Correct option: (b)

Mode = (3 x median) – (2 x mean)

**Question 13.**

**Solution:
**Correct option: (c)

Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5

**Question 14.**

**Solution:
**

**Question 15.**

**Solution:
**Correct option: (c)

Class having maximum frequency is the modal class.

Here, maximum frequency = 30

Hence, the modal class is 30 – 40.

**Question 16.**

**Solution:
**Correct option: (b)

Mode = \({ x }_{ k }+h.\left\{ \frac { { f }_{ k }-{ f }_{ k-1 } }{ \left( { 2f }_{ k }-{ f }_{ k-1 }-{ f }_{ k+1 } \right) } \right\} \)

**Question 17.**

**Solution:
**Correct option: (a)

Median = \(l+\left\{ h\times \frac { \left( \frac { N }{ 2 } -cf \right) }{ f } \right\} \)

**Question 18.**

**Solution:
**Correct option: (c)

Mean = 8.9

Median = 9

Mode = 3Median – 2Mean

= 3 x 9 – 2 x 8.9

= 27 – 17.8

= 9.2.

**Question 19.
**

**Solution:**

**Question 20.
**

**Solution:**

**Question 21.**

**Solution:
**

**Question 22.**

**Solution:
**

**Question 23.**

**Solution:
**Correct option: (c)

For a symmetrical distribution,

we have Mean = mode = median

**Question 24.
**

**Solution:**

**Correct option: (c)**

Number of families having income more than Rs. 20000 = 50

Number of families having income more than Rs. 25000 = 37

Hence, number of families having income range 20000 to 25000 = 50 – 37 = 13

**Question 25.**

**Solution:
**

**Question 26.**

**Solution:
**Correct option: (d)

Mean of 20 numbers = 0

Hence, sum of 20 numbers = 0 x 20 = 0

Now, the mean can be zero if

sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),

sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S), …….

sum of 19 numbers is (S) and the 20

^{th}number is (-S), then their sum is zero.

So, at the most, 19 numbers can be greater than zero.

**Question 27.**

**Solution:
**

**Question 28.**

**Solution:
**

**Question 29.**

Solution:

**(a) – (s)
**The most frequent value in a data is known as mode.

**(b) – (r)**

Mean cannot be determined graphically.

**(c) – (q)**

An ogive is used to determine median.

**(d) – (p)**

Standard deviation is not a measure of central tendency.

**Question 30.**

**Solution:
**

**Question 31.
**

**Solution:**

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