CBSE 10th Standard Maths Subject Statistics Ncert Exemplar 4 Marks Questions With Solution 2021
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CBSE 10th Standard Maths Subject Statistics Ncert Exemplar 4 Marks Questions With Solution 2021
10th Standard CBSE
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Reg.No. :
Maths
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Calculate the mean of the scores of 20 students in a Mathematics test.
Marks 10-20 20-30 30-40 40-50 50-60 Number of students 2 4 7 6 1 (a) -
Determine the mean of the following distribution.
Marks Number of students Below 10 5 Below 20 9 Below 30 17 Below 40 29 Below 50 45 Below 60 60 Below 70 70 Below 80 78 Below 90 83 Below 100 85 (a) -
An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table:
Number of seats 100-104 104-108 108-112 112-116 116-120 Frequency 15 20 32 18 15 Determine the mean number of seats occupied over the flights.
(a) -
The mileage (in km/L) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below:
Mileage (in km/L) 10-12 12-14 14-16 16-18 Number of cars 7 12 18 13 Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km/L. Do you agree with this claim?
(a) -
The mean of the following frequency distribution is 50, but the frequencies f1 and f2 in classes 20-40 and 60-80 respectively are missing. Find the missing frequencies.
Class interval 0-20 20-40 40-60 60-80 80-100 Total Frequency 17 f1 32 f2 19 120 (a)
4 Marks
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CBSE 10th Standard Maths Subject Statistics Ncert Exemplar 4 Marks Questions With Solution 2021 Answer Keys
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35
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Here, we observe that 5 students have scored marks below 10, i.e it lies between class interval 0-10 and 9 students have scored marks below 20.
So, (9-5)=4 students lie in the class interval 10-20. Continuing in the same manner, we get the following frequency distribution table for given data.Marks Number of students (fi) Class marks (xi) \(=\frac { x_{ i }^{ u_{ i } }-45 }{ h } \) fiui 0-10 5 5 -4 -20 10-20 9-5=4 15 -3 -12 20-30 17-9=8 25 -2 -16 30-40 29-17=12 35 -1 -12 40-50 45-29=16 45 0 0 50-60 60-45=15 55 1 15 60-70 70-60=10 65 2 20 70-80 78-70=8 75 3 24 80-90 83-78=5 85 4 20 90-100 85-83=2 95 5 10 Total \(n=\sum { f_{ i } } =85\) \(\sum { f_{ i }u_{ i } } =29\) Here, assumed mean(a)=45 and class width (h)=10
By step deviation method,
Mean \(\left( \overline { x } \right) =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h\)
\(=45+\left\{ \frac { 29 }{ 85 } \right\} \times 10=45+\frac { 58 }{ 17 } \\ =45+3.41=48.41\) -
109.92
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14.48km/L No, the manufacturer is claiming mileage 1.52km/L more than the average mileage.
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f1=28, f2=24
4 Marks