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10th Grade > Mathematics

STATISTICS MCQs

Total Questions : 60 | Page 5 of 6 pages
Question 41.


Class mark of a class is _____


  1.     Upper limit + lower limit
  2.     upper limit + lower limit2
  3.     upper limit - lower limit2
  4.     upper limit - lower limit
 Discuss Question
Answer: Option B. -> upper limit + lower limit2
:
B

Class mark is the midpoint of a class interval.
Therefore, its formula is given by upper limit + lower limit2.


Question 42.


The mean of the following distribution is ____.
xi10131619fi2576


  1.     15.2
  2.     15.35
  3.     15.55
  4.     16
 Discuss Question
Answer: Option C. -> 15.55
:
C

xififixi1022013565167112196114fi=20 fixi=311


Mean =fixifi
         
          =31120 
         
          =15.55


Question 43.


If the mean of the following data is 17 then the value of p is 


___


xi10p182125fi1015799


 Discuss Question
Answer: Option C. -> 15.55
:

xififixi1010100p1515p187126219189259225
fi=50
fixi = 640+15p
Given mean=17
17 = 640+15p50
Solving, we get p = 14


Question 44.


If the median of the following data is 50, then find p.
ClassFrequency0 - 201020 - 40740  - 601660 - 80p80 - 1008


  1.     6
  2.     7
  3.     8
  4.     9
 Discuss Question
Answer: Option D. -> 9
:
D

 Class FrequencyCumulativeFrequency0 - 20101020 - 4071740 - 60163360 - 80p33+p80 -100841+p


 


 Since median is 50, median class is 40-60.
Using
Median=l+[n2cff]×h,
Here,n=41+p,
 n2=(41+p)2
Now, 50=40+[(41+p)21716×20]
10=[(41+p34)216×20]
10=[(7+p)16×2×20]
Simplifying, we get (7+p)16=1
7+p=16
p=9


Question 45.


Mode is


  1.     Most frequent value
  2.     Least frequent value
  3.     Middle most value
  4.     Average value
 Discuss Question
Answer: Option A. -> Most frequent value
:
A

Mode is the most frequently observed value


Question 46.


If the preceding and succeeding classes of the modal class have the same frequency, then the mode will be at the midpoint of the modal class.


  1.     True
  2.     False
  3.     Frequency of the class succeeding the modal class
  4.     Frequency of median class
 Discuss Question
Answer: Option A. -> True
:
A

The formula for the mode is as follows.


Mode=l+(f1f02f1f0f2)×h
 
l= lower boundary of the modal class
h= size of the modal class interval
f1=  frequency of the modal class.
f0=  frequency of the class preceding the modal class
f2= frequency of the class succeeding the modal class



If the preceding and succeeding classes have the same frequency, then f0=f2=f(say).


Then the equation reduces to
Mode=l+(f1f2f1ff)×h
Mode=l+(f1f2(f1f))×h
Mode=l+12×h, which is the midpoint of the modal class.


 If the preceding and succeeding classes of the modal class have the same frequency, then the mode will be at the midpoint of the modal class.


Question 47.


In the fromula for mode of a grouped data ,
mode =l+[t1t02f1f0f2]×h ,
where symbols have their usual meaning f1   represents 


  1.     Frequency of Modal class
  2.     Frequency of the class preceding the modal class
  3.     Frequency of the class succeeding the modal class
  4.     Frequency of median class
 Discuss Question
Answer: Option A. -> Frequency of Modal class
:
A

In the formula for mode, f1 represents the frequency of the modal class.


Question 48.


The class mark for a class interval while calculating the mean is its upper limit.


  1.     True
  2.     False
  3.     Frequency of the class succeeding the modal class
  4.     Frequency of median class
 Discuss Question
Answer: Option B. -> False
:
B

The class mark is the midpoint of the lower and upper limits of a class. It is calculated by taking the average of the upper and lower limits i.e.
Class mark = Upper limit+Lower limit2


Question 49.


fx = 5x + 2, f = 12. If the mean of the distribution is 6, what is the value of x?


  1.     12
  2.     14
  3.     15
  4.     18
 Discuss Question
Answer: Option B. -> 14
:
B

We know Mean = fxf


Hence, 5x+212 = 6 


5x + 2 = 12 × 6
5x = 72 – 2 = 70
x = 14


Question 50.


Use the empirical relationship and find the value of mode when the value of median and mean are 6 and 7 respectively.


  1.     4
  2.     3
  3.     6
  4.     7
 Discuss Question
Answer: Option A. -> 4
:
A
We know that,
3 Median = Mode + 2 Mean 
or 
Mode = 3 Median - 2 Mean 
          =  3×62×7
          = 18 - 14
          = 4 
Therefore, mode = 4 

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