[From (i)
⇒ 3f1 + 329 7f1 = 257
⇒ -4f1 = 257 – 329 = -72
⇒ f1 = \(\frac{-72}{-4}\) = 18
Putting the value of f1 in (i), we get
f2 = 47 f1
⇒ f2 = 47 18 = 29
∴ f1 = 18, f2 = 29Question 32.com.
The distribution below gives the weights of 30 students of a class. The study material for Class 10 for Real Numbers has been made by experienced teachers of leading schools in India is available for free download in pdf Central Academy for Police Training, (CAPT), Bhopal, Bureau of Police Research Development, Ministry of Home Affairs, Government of India, is offering an online Student Outreach program on “Road Safety: Roles Responsibility” for students internet Class IX to.

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5x
⇒ 36 = 18x ⇒ x = 2
From (i), y = 4 x = 4 2 = 2 ∴ x = continue reading this y = 2Question 34. Also, download the BYJU’S App for interactive study videos.Revise all syllabus before solving WorksheetsA student must revise the entire Syllabus for Class 10 Mathematics before solving worksheets in order to get the best performance out of themselves. But the frequencies f1 and f2 in class 20 40 and 60 80 respectively are missing.

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Find the missing frequencies: (2013)

Solution:

⇒ 2730 + 30f1 + 70f2 = 5300
⇒ 30f1 + 70f2 = 5300 2730 = 2570
⇒ 3f1 + 7f2 = 257 [Dividing by 10
⇒ 3f1 +7(47 f1) = 257 ..

Solution:
Here, we use step deviation method to find mean.
Find the median class of the following distribution:

Solution:
First we find the cumulative frequency

Here, \(\frac{n}{2}\) = \(\frac{50}{2}\)
∴ Median class = 30 40.
If xis are the mid-points of the class intervals of a grouped data. Term I.

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So 35 40 is the median class.. Find the value of P: (2012, 2017D)

Solution:
Question 7. Carefully understand the syllabus for Class 10 Statistics and download the worksheets for the topics which you have studied today.

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Students can download free printable worksheets for practice, topic wise solved questions for all topics and chapters.
Find the mean of the first five natural numbers. You can download free printable assignments for practice, topic-wise question banks for all chapters.__mirage2 = {petok:”2530925766f7324fb3fffe5289d9d27ab6a2d09a-1664042102-1800″};
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Solution:

∴ Median class 1700 1850.

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Step 3: Median = l + [(N/2 cf)/f] × h
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). So, CBSE Class 10 students are advised to memorise these questions thoroughly. Find the mean literacy rate: (2012)

Solution:
Question 17.The marks scored by 100 students out of 50 marks are given below.

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Ungrouped data is data in its original or raw form.
Solution:

Hence, median marks = 24Question 5.

Solution:
Calculation of median

The cumulative frequency just greater than \(\frac{n}{2}\) = 15 is 19, and the corresponding class is 55 60.To know more about Median, visit here.967)Mean = 47., the mean of nth and (n + 1)th term will be the median.

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Start with the Highest marks weight-age topicsAccording to the examination pattern and Sample Papers for Mathematics provided by CBSE, all topics are divided on the basis article marks weight-age.Step 4: Find the deviation of ′a′ from each of the x′is
di=xi−aStep 5: Find the mean of the deviationsStep 6:  Calculate the mean as
For example, let us consider the same example as provided above.Now, using the assumed mean method formula, we can getMean = 47.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Term I and Term II.

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Hence, modal lifetime of the components is 65.Step 4: Find the deviation of a from each of the x′is
di=xi−aStep 5: Divide all deviations −di by the class width (h) to get u′is.5 from the lower limits. Access free CBSE NCERT printable assignments for Class 10 Real Numbers with solutions prepared by expert teachers.

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The study material for Class 10 for Statistics has been made by experienced teachers of leading schools in India is available for free download in pdf Ministry of Education, Govt of India vide letter No… (2013)

Solution:

∴ Frequency of class 30 40 = 3Question 4.5, cf = 14, f = 14, h = 2

Hence, median is verified. Students will be able to identify all mistakes in understanding of the topics.

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Solution:
We are given the cumulative frequency distribution. During this camp, a medical check of 35 students was done and their weights were recorded as follows: (2016)

Compute the modal weight. The study schedule can be prepared by dividing all syllabus for Class 10 Statistics on the basis of their mark distribution, along with the allotment of maximum time to the most important topics.Question 5.

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, xn are the given observations and their respective frequencies are f1, f2, . This year to avoid any confusion and conflict, CBSE has decided to reduce the syllabus into term 1 and term 2., fn respectively, then the mean of the grouped data (with class interval) is given as follows:Where fi is the frequency of ith class whose class mark is xi  And, i varies from 1 to n. The formula is valid for equal class intervals and when the modal class is unique.

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So, the modal class is 60 80.Question 8.
Draw: (a) more than ogive and (2017OD)
(b) less than ogive for the following data
Also find its median. Find the median of the given data:Now, arrange the marks scored by students in the ascending order.77Therefore,Mean = 137.

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The mean of the following frequency distribution is 62..

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