Find the median of the following distribution Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency 3 6 8 15 10 8

The cumulative frequency table is as below, Class interval Frequency f i Cumulative frequency c . f . 0 - 10 3 3 10 - 20 6 6 + 3 = 9 20 - 30 8 8 + 9 = 17 30 - 40 15 15 + 17 = 32 40 - 50 10 10 + 32 = 42 50 - 60 8 8 + 42 = 50 Total N = ∑ f i = 50 We need to find the median of the given frequency distribution table. We have, N = 50 So, N 2 = 50 2 = 25 The cumulative frequency just greater than 25 is 32 , which corresponds to the class 30 - 40 . Thus, the median class is 30 - 40 . Here, l = 30 (the lower limit of the median class) f = 15 (frequency of the median class) h = 10 (width of each class) N = 50 (number of observations) and C = 17 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h = 30 + 25 - 17 15 × 10 = 30 + 8 15 × 10 = 30 + 16 3 = 30 + 5 . 33 = 35 . 33 approx Hence, the median is 35 . 33 .

A survey was conducted in a class to find the preference of students for the soft drinks and recorded as follows: Soft drinks Number of students Orange juice Cola Lime juice Mango juice Melon Juice 15 13 9 7 11 Make a bar graph with a suitable scale for the above data.

A survey was conducted in a class to find the preference of students for the soft drinks and recorded as follows: Soft drinks Number of students Orange juice Cola Lime juice Mango juice Melon Juice 15 13 9 7 11 The bar graph is drawn by taking the soft drinks on horizontal axis and number of students on vertical axis. The bar graph for the above data is shown below:

Define range of a grouped data.

In the case of continuous frequency distribution or grouped data, the range is defined as the difference between the upper limit of the maximum interval of the grouped data and the lower limit of the minimum interval. It is the simplest measure of dispersion. It gives a comprehensive view of the total spread of the observations. Thus, the formula to calculate the range of a grouped data is given below: Range = Upper-class boundary of the highest interval – Lower class boundary of the lowest interval

Find the median of the following distribution Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency 3 6 8 15 10 8

The cumulative frequency table is as below, Class interval Frequency f i Cumulative frequency c . f . 0 - 10 3 3 10 - 20 6 6 + 3 = 9 20 - 30 8 8 + 9 = 17 30 - 40 15 15 + 17 = 32 40 - 50 10 10 + 32 = 42 50 - 60 8 8 + 42 = 50 Total N = ∑ f i = 50 We need to find the median of the given frequency distribution table. We have, N = 50 So, N 2 = 50 2 = 25 The cumulative frequency just greater than 25 is 32 , which corresponds to the class 30 - 40 . Thus, the median class is 30 - 40 . Here, l = 30 (the lower limit of the median class) f = 15 (frequency of the median class) h = 10 (width of each class) N = 50 (number of observations) and C = 17 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h = 30 + 25 - 17 15 × 10 = 30 + 8 15 × 10 = 30 + 16 3 = 30 + 5 . 33 = 35 . 33 approx Hence, the median is 35 . 33 .

Define class-mark of a class interval.

We are given to define the class-mark of a class interval. When the data in statistics is expressed in the form of a group or class, it is always associated with the upper-class limit and the lower class limit. The class-mark of a class interval is defined as the average of the lower and upper-class limits. For example, if p is the lower class limit and q is the upper-class limit of a class, then the class-mark is given by p + q 2 . Enable Ginger Cannot connect to Ginger Check your internet connection or reload the browser Disable in this text field Rephrase Rephrase current sentence Edit in Ginger ×

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>meghalaya board-2019>Q 6

MEDIUM

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IMPORTANT

Earn 100

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

Daily pocket allowance (in $₹$ ) $11 - 13$ $13 - 15$ $15 - 17$ $17 - 19$ $19 - 21$ $21 - 23$ $23 - 25$

Frequency $7$ $6$ $9$ $13$ $f$ $5$ $4$

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Important Questions on Statistics

HARD

10th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>meghalaya board-2020>Q 8

The mean of the following frequency distribution is $62.8$ and the sum of all frequencies is $50$ . Find the missing frequencies $f_{1}$ and $f_{2}$ :