Question

The marks obtained by 70 students in an examination are grouped as
follows:
Marks
0-10 10-20 20-30 | 30-40 40-50 50-60
Number of Students
8 10 25
7. 16
Calculate:
(1) The marks at the 25th percentile.
(ii) The marks at the third quartile.
(iii) The approximate number of students who failed in the examinations if
the pass marks are considered to be 40%.​

Answer 1

ANSWER

Marks                     No. of students(f)       Cumulative Frequency0−105510−2091420−30163030−40225240−50267850−60189660−701110770−80611380−90411790−1003120n=120On the graph paper, we plot the following points:

(10,5),(20,14),(30,30),(40,52),(50,78),(60,96),(70,107),(80,113),(90,117),(100,120).

(i) Median =(2n)th term [∵n=120, even]

=2120=60th term

From the graph 60th term =42

(ii) The number of students who obtained more than 75% marks in test

=120−110

=10.

(iii) The number of students who did not pass the test if the minimum pass marks 

Answer 2

given table of number of students and marks obtains, answer the question based on the given information

Explanation:

1. total number of students $N=70$  , marks obtained by them are,                              $Marks \ \ \ \ \ No.\ of\ students \ \ \ \ \ \ cf\\0-10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8\\10-20\ \ \ \ \ \ \ \ \ \ \ \ \ 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 18\\20-30\ \ \ \ \ \ \ \ \ \ \ \ \ 25\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 43\\30-40\ \ \ \ \ \ \ \ \ \ \ \ \ 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 50\\40-50\ \ \ \ \ \ \ \ \ \ \ \ \ 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 66\\50-60\ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 70$        
2. hence, nth percentile (P) of given total number of values (T) is given by,                           $P=\frac{n}{100}T$                                                                                                                    here, we need to find $25^{th$ percentile -> $n=25\ \ \ \ T=60$                                             hence, $P_{25^{th}}=\frac{25}{100}(60) = 15$                   ---->ANSWER (i)
3. third quartile of a grouped data is given by ,  $Q_3=l+(\frac{\frac{3N}{4}-cf }{f} )i$    --(i)                      here, l- lower limit of third quartile class ,N-total frequency, cf- cumulative frequency before quartile class, f- frequency of quartile class, i- width of quartile class          
4. first to find third quartile class, $\frac{3N}{4}= \frac{3(70)}{4}= 52.5$    hence third quartile class is $(30-40)$ as $52.5$ lies between $cf-50,66$
5. now, we have $l=30,\ cf=43,\ f=7,\ i=11$ putting these in (i),                                                           $Q_3=30+(\frac{52.5-43 }{7} )11\\Q_3= 45(approx.)$   ------ANSWER (ii)  
6. total marks are $60$ hence $40\%$ are, $60(\frac{40}{100})=24$                                                  hence number students having marks less than $24$  are failing on the examination