Question

9. The following table shows the Marks obtained by 100 candidates in an examination. Calculate the
mean, median and standard deviation.
Marks
1-10
11-20
21-30
31-40
41-50
51-60
No. of candidates
3
16
26
31
16
S.​

Answer 1

$ANSWER</p><p></p><p>Here we have, the cumulative frequency distribution. So, first we convert it into an ordinary frequency distribution. we observe that are 80 students getting marks greater than or equal to 0 and 77 students have secured 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3.</p><p>Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the following frequency distribution.</p><p></p><p>Marks Number of students0-10310-20520-30730-401040-501250-601560-70 1270-80680-90290-1008Now, we compute mean arithmetic mean by taking 55 as the assumed mean.</p><p>Computative of Mean </p><p>Marks</p><p>(xi)Mid-value (fi)Frequency ui10xi−55fiui0-1053-5-1510-20155-4-2020-30257-3-2130-403510-2-2040-504512-1-2050-6055150060-70651211270-8075621280-908523690-100958432Total          </p><p>       ∑fi=80∑fiui= -26</p><p> We have,</p><p>N= sumfi=80,∑fiui=−26, A= 55 and h= 10 </p><p>∴X=A+h[N1∑fiui]</p><p>⇒X=A+h[N1∑fiui]</p><p>⇒X=55+10×80−26=55−3.25=51.75Marks. </p><p></p><p>$

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Answer 2

Answer:

Step-by-step explanation: