- Flash back from IX Class
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- Mean for Ungroup Frequency table
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- Mean for group Frequency table
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- Various ways to calculate mean
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- Mode for grouped frequency table
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- Cumulative Frequency chart
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- Median of a grouped data frequency table
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- Empirical Formula between Mode, Mean and Median

Given below are the

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

(b) The sum of deviations of a set of values {a,b,c,d,e,f,g.h.i……} n items measured from 26 is -10 and the sum of deviations of the values from 20 is 50. The value of n is …(10/12/9) And mean of the items is ……. ( 19/18/17/25)

(c) For a given data with 110 observations the ‘less than ogive’ and the ‘more then ogive’ intersect at (18, 20). The median of the data is …… (18/20/19)

(d) The curve drawn by taking upper limits along x-axis and cumulative frequency along y-axis is ……….( less than ogive /more than ogive)

(e) The mean of five numbers is 40. If one number is excluded, their mean becomes

28. The excluded number is……. (68 / 88)

(f) The modal class of the grouped size frequency table given below is

5-5.2 |
5.2-5.4 |
5.4-5.6 |
5.6-5.8 |
5.8-6.0 |

34 |
4 |
4 |
4 |
6 |

(a) 24

So M=22

(b) 10,25

Subtracting ,we get

-6n=-60 => n=10

So $\sum x=250$

- Mean=250/10=25

(d) less than ogive curve

(e) 88

Sum of five number=5Xmean=200

Sum of four number=4Xmean=112

Subtracting, we get the number=88

f) Modal class is 5-5.2

Class interval |
5-15 |
15-25 |
25-35 |
35-45 |
45-55 |
55-65 |

Frequency |
6 |
11 |
21 |
23 |
14 |
4 |

b) The modal class is 35-45

c) The mode is 34

d) The Frequency for less than is 35 is 38

e) The median is

Class interval |
5-15 |
15-25 |
25-35 |
35-45 |
45-55 |
55-65 |

Frequency |
6 |
11 |
21 |
23 |
14 |
5 |

Class mark |
10 |
20 |
30 |
40 |
50 |
60 |

F_{i}x_{i} |
60 |
220 |
630 |
920 |
700 |
300 |

Where

l = lower limit of the modal class,

h = size of the class interval (assuming all class sizes to be equal),

f

f

f

Now l=35 h=10 f

So substituting all the values we get

M

Now

3 Median=Mode +2 Mean

So median =Mode +2 Mean)/3=35.853

a) False. I

b) True

c) False

d) True. e) False

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

Which of the following is the measure of central tendency?

a) Mean

b)Mode

c) Median

d) Range

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

Find the mean of 32 numbers given mean of ten 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

Mean of 10 number =12

- Sum of these 10 numbers =120

- Sum of these 20 numbers =180

- Sum of these 2 numbers =20

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

Mean =5.5

Median is mean of 5 and 6 th term, So 5

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

The following observations are arranged in ascending order :

20, 23, 42, 53, x, x + 2, 70, 75, 82, 96

If the median is 63, find the value of

a) 62

b) 64

c) 60

d) None of these

Median is mean of 5 and 6 term

So x+1=63

X=62

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

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

Then

X+25=20*60=1200

Now

X+125=20*y=20y

Subtracting

125-25=20y-1200

20y=1300

y=65

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 |

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 |
5 |

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

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

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 l=130 n=266 cf=89 f=72 h=10

Substituting these

M

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 |

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