# NCERT Class 10 Mathematics Statistics Formative Assessment CBSE Board Sample Problems Part 2 (For CBSE, ICSE, IAS, NET, NRA 2023)

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**Multiple-choice Questions**

3) While computing mean of grouped data, we assume that the frequencies are

a) Centred at the upper limits of the classes

b) Centred at the lower limits of the classes

c) Centred at the class-marks of the classes

d) Evenly distributed over all the classes

Solution (c)

Question 4

Which of the following is the measure of central tendency?

a) Mean

b) Mode

c) Median

d) Range

Solution (a) , (b) , (c)

Question 5

For drawing a frequency polygon of a continuous frequency distribution, we plot the Points whose ordinates are the frequencies of the respective classes and abscissae are respectively:

a) Class marks of the classes

b) Upper limits of preceding classes

c) Lower limits of the classes

d) Upper limits of the classes

Solution (a)

Question 6

Find the mean of 32 numbers given mean often of them is 12 and the mean of other 20 is 9. And last 2 number is 10

a) 10

b) 12

c) 13

d) 14

Solution (a)

Mean of 10 number = 12

Sum of these 10 numbers = 120

Mean of 20 number = 9

Sum of these 20 numbers = 180

Mean of 2 number = 10

Sum of these 2 numbers = 20

Mean of 32 number

Question 7

The median and mean of the first 10 natural numbers.

a) 5.5,5.5

b) 5.5,6

c) 5,6

d) None of these

Solution (a)

Mean = 5.5

Median is mean of 5 and 6^{th} term, So 5

Question 8

Anand says that the median of 3, 14, 19,20, 11 is 19. What doesn՚t the Anand understand about finding the median?

a) The dataset should be ascending order

b) Highest no in the dataset is the median

c) Average of lowest and highest is the median

d) None of these

Solution (a)

Question 9

The following observations are arranged in ascending order:

If the median is 63, find the value of x.

a) 62

b) 64

c) 60

d) None of these

Solution (a)

Median is mean of 5 and 6 term

So

Question 10

The mean of 20 observations was 60. It was detected on rechecking that the value of 125 was wrongly copied as 25 for computation of mean. Find the correct mean

a) 67

b) 66

c) 65

d) None of the above

Solution

Let x be the sum of observation of 19 numbers leaving 125,

Then

Now

Subtracting

Question 11

Compute the Median for the given data

Class interval | 100 - 110 | 110 - 120 | 120 - 130 | 130 - 140 | 140 - 150 | 150 - 160 |

Frequency | 6 | 35 | 48 | 72 | 100 | 4 |

Solution

First, calculate the Class mark and cumulative frequency of the data

Class interval | 100 - 110 | 110 - 120 | 120 - 130 | 130 - 140 | 140 - 150 | 150 - 160 |

Frequency | 6 | 35 | 48 | 72 | 100 | 4 |

Class Mark | 105 | 115 | 125 | 135 | 145 | 155 |

Cumulative Frequency | 6 | 41 | 89 | 161 | 261 | 266 |

We have N = 266, So , Cumulative frequency first greater than 133, lies in class 130 - 140

So median class is 130 - 140

Now

Median is calculated as

Where

l = lower limit of median class,

n = number of observations,

cf = cumulative frequency of class preceding the median class,

f = frequency of median class,

h = class size (assuming class size to be equal)

Here

Substituting these

**Match the column**

A histogram | Is the diagram showing a system of connections |

Or interrelations between two or more things by using bars | |

Discontinuous Frequency Distribution | A frequency distribution in which the upper limit of one class coincides from the lower limit of the succeeding class |

Continuous Frequency Distribution | Is the bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval. |

Ogive is the graph of | Is a set of adjacent rectangles whose areas are proportional to the frequencies of a given continuous frequency distribution? |

lower/upper limits and cumulative frequency |