CBSE 9th Mathematics - Statistics Four Marks Questions | QB365

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CBSE 9th Mathematics - Statistics Four Marks Questions

Published on: 09/10/2019

Statistics

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Std: 9th Standard CBSE - Mathematics

chapter: - Model Question paper

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Questions:

Generic 4

10 x 4 = 40

The mean of 10,12,16,20,p and 26 is 17. Find the value of p.

The mean of 10,12,18,13,x and 17 is 15. Find the value of x.

Calculate mean of first 5 prime numbers.

Find the mean of first 10 multiples of 7?

Draw a histogram and frequency polygon of the following data:

Marks	No.of students
20-30	5
30-40	12
40-50	6
50-60	20
60-70	18
70-80	10
80-90	16
90-100	3

Draw a frequency polygon to represent the following information:

Class	Frequency
25-29	5
30-34	15
35-39	23
40-44	20
45-49	10
50-54	7

The following table gives the performance of 90 students in a mathematics test of 100 marks.

Marks	Number of students
0-20	07
20-30	10
30-40	10
40-50	20
50-60	20
60-70	15
70-above	08
Total	90

Represent the given information with the help of a histogram.

The marks scored by 750 students in examination are given in the form a frequency distribution table.

Marks	No. of students
600-640	16
640-680	45
680-720	156
720-760	284
760-800	172
800-840	59
840-880	18

Draw a histogram to represent the above data.

The following data gives the number (in thousands) of applicants registered with an Employment Exchange during 2005-2010.

Year	No.of applications registered (in thousands)
2005	19
2006	21
2007	23
2008	30
2009	32
2010	36

Construct a bar graph to represent the above data.

10.

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. Find the class size and class intervals.

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Answers Key:

Generic 4

10 x 4 = 40

\(\frac{10+12+16+20+p+26}{6}=17\)
\(\Rightarrow\) 84 + p = 102
\(\Rightarrow\) p = 102 - 84
\(\Rightarrow\) p = 18.

According to the question,
\(\frac {10+12+18+13+x+17}{6}=15\)
\(\Rightarrow\) x + 70 = 90
\(\Rightarrow\) x = 20

First 5 prime numbers are 2,3,5,7,11
Their mean \(= \frac {2+3+5+7+11}{5}=\frac{28}{5}=5.6\)

First 10 multiples of 7 are
7,14,21,28,35,42,49,56,63,70
Their mean
\(\frac {7+14+21+28+35+42+49+56+63+70}{10} = \frac {385}{10}=38.5\)

We shall first make the class intervals continuous. Then, the modified table is as follows:

Class	Class-marks	Frequency
24.5-29.5	27	5
29.5-34.5	32	15
34.5-39.5	37	23
39.5-44.5	42	20
44.5-49.5	47	10
49.5-54.5	52	7
Total		80

Adjusted frequency a class \(\frac {Minimum \ class \ size}{Class \ size}\times {Frequency \ of \ the \ class}\)
Minimum class size = 10

Marks	Number of students	Adjusted frequency
0-20	07	\(\frac {10}{20}\times 7 = 3.5\)
20-30	10	\(\frac {10}{10}\times 10 = 10\)
30-40	10	\(\frac {10}{10}\times 10 = 10\)
40-50	20	\(\frac {10}{10}\times 20 = 20\)
50-60	20	\(\frac {10}{10}\times 20 = 20\)
60-70	15	\(\frac {10}{10}\times 15 = 15\)
70-80	08	\(\frac {10}{10}\times 08 = 08\)

10.

Class size = 114 -104 = 10
\(\frac {10}{2}=5\)
The class intervals are
104-5___104+5, i.e., 99-109;
114-5___ 114+5, i.e, 109-119;
124-5___ 124+5, i.e., 119-129;
134-5___ 134+5, i.e., 129-139;
144-5___ 144+5, i.e., 139-149;
154-5___ 154+5, i.e., 149-159;
164-5___ 159-169.