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801.

Can mode be calculated for grouped data with unequal class sizes?

Answer»

Yes. Mode can be calculated for grouped data with unequal class sizes.
Discussion
802.

The mean for a grouped data is calculated by \(\bar{x}\) = a + Σfidi  / Σfi What do the terms ‘f.’ and ’d.’ represent in the above formula .

Answer»

fi = frequency of the ith observation 

di = deviation of the ith observation
Discussion
803.

The mean of a + 1,a + 3,a + 4 and a + 8 isA) a + 7B) a + 4 C) a – 3 D) None

Answer»

Correct option is: B) a + 4

Mean = \(\frac {sum \, of\, obs.}{Number \, of\, obs.}\) = \(\frac {(a+1)+ (a+3) + (a+4) + (a+8)}{4} \)

\(= \frac {4a+16}4 = \frac {4(a+4)}{4} = a + 4\)

Correct option is: C) a + 4
Discussion
804.

In the formula of Mode in the grouped data ‘l’ represents?A) lower limit of the class with highest frequencyB) lower limit of the class with lowest frequency C) upper boundary of the class with highest frequency D) lower boundary of the class with low frequency

Answer»

Correct option is: A) lower limit of the class with highest frequency

Mode for grouped data is given by

Mode = \(l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h\) 

Where l = lower limit of the class with highest frequency.

Correct option is: A) lower limit of the class with highest frequency
Discussion
805.

Find the median of 7,6,5,3,9,4,3.

Answer» Arranging the data in ascending order we get 3,3,4,5,6,7,9
The number of observations `n = 7` which is odd
`therefore " "Median"" = ((n+1)/(2))th " observations"`
`= ((7+1)/(2))th" "observations"`
`=4th" observations"`
`=5`
Discussion
806.

The formula of mode isA. `L+[(f_1-f_0)/(2f_1-f_0-f_2)]xxh`B. `L-[(f_1-f_0)/(2f_1-f_0-f_2)]xxh`C. `L+[(f_1-f_0)/(2f_1-f_0+f_2)]xxh`D. `L-[(f_1-f_0)/(2f_1-f_0+f_2)]xx1/h`

Answer» Correct Answer - A
Discussion
807.

Find the median of the following data 3,5,9,10,11,4,5,8,12,15

Answer» On arranging the given data in ascending order, we get 3,4,5,5,8,9,10,11,12,15
Hence, no. of terms `(n) = 10` which is an even number
`"Therefore, median " = (((n)/(2))th" "term + ((n)/(2)+1)th" "term")/(2)`
`= (((10)/(2))th" "term + ((10)/(2)+1)th" "term")/(2)`
`= (5th" "term + 6th" "term")/(2) = (8+9)/(2) = 8.5`
Discussion
808.

The mode of the distribution 3,5,7,4,2,1,4,3,4 isA. `7`B. `4`C. `3`D. `1`

Answer» Correct Answer - B
Discussion
809.

Find the mode of 4,6,2,2,1,3,7,9,2,3,2.

Answer» Here, frequency of 2 is highest i.e., 2 occur most times.
So, the mode `=2.`
Discussion
810.

Find the mode of the following frequency table, which gives the marks scored by 40 students in a test:

Answer» The marks obtained 5 has highest frequency. Therefore, mode is 5.
Discussion
811.

The average scored by the students of a class in English is 64. The average of marks scored by boys and the girls are respectively 68 and 58. Then find the ratio of the number of boys to the number of girls.

Answer» Let `n_(1)` boys and `n_(2)` girls are there in a class
Average marks of boys `=68`
`rArr "Sum of marks of all the boys" = 68n_(1) " " (sumx_(i) = n bar(x))`
Similarly,
Sum of marks of all the girls `=58n_(2) " " (sumx_(i) = n bar(x))`
`therefore "Sum of marks of all students " = 68n_(1) + 58n_(2)`
`therefore " Average marks" = (68n_(1) + 58n_(2))/(n_(1)+n_(2))= 64 " " `(given)
`rArr 68n_(1) + 58n_(2) = 64n_(1)+64n_(2)`
`rArr 4n_(1) = 6n_(2) rArr (n_(1))/(n_(2))=(6)/(4)=(3)/(2)`
`therefore "Required ratio " = 3:2`
Discussion
812.

In an examination, the mean of marks scored by a class of 30 students was calculated as 58.5 Later on, it was detected that the marks of one student was wrongly copied as 57 instead of 75. Find the correct mean.

Answer» `"Mean of marks" = ("incorrect marks of 30 students")/(30)`
`therefore" "58.5 = ("incorrect sum of marks")/(30)`
` rArr " ""Incorrect sum of marks" = 58.5 xx 30 = 1755`
As marks of one student was wrongly copied as 57 instead of 75, correct sum of marks
`="Wrong sum - Wrong value + Correct value"`
`=1755 - 57 + 75 = 1773`
`therefore" ""Correct mean " = (1773)/(30) = 59.1`
Discussion
813.

Find the missing value of p for the following distribution whose mean is 12.58x:581012p2025y:25822742

Answer»
Xyyx
5210
8540
10880
1222264
p77p
20480
25250
N = 50\(\sum\)yx = 524 + 7p

Given,

Mean = 12.58

\(\frac{\sum yx}{N}=12.58\)

\(\frac{524+7p}{50}=12.58\)

524 + 7p = 12.58 (50)

7p = 629 – 524

p = \(\frac{105}{7}=15\)
Discussion
814.

The standard deviation of 25 numbers is 40. If each of the numbers in increased by 5, then the new standerd deviation will be -A. 40B. 45C. `40+(21)/(25)`D. None of these

Answer» Correct Answer - A
If each item of the data is increased or decreased by the same constant, the standard deviation of the data remains unchanged.
Discussion
815.

Find the quartile deviation of the following discrete series. `{:(x,3,5,6,8,10,12),(f,7,2,3,4,5,6):}`A. 4B. 3C. 3.5D. 4.5

Answer» Correct Answer - C
(i) `QD = (Q_(3) - Q_(1))/(2)`
(ii) Write cumulative frequency.
(iii) The corresponding values of x of `(N)/(4)`th and `(3N)/(4)`th observation is `Q_(1) "and" Q_(3)` respectively.
Discussion
816.

What is Cumulative frequency curve ( an Ogive curve ). How many methods are to construct ogives? Explain.

Answer»

Cumulative frequency curve or an Ogive curve : The graphical representation of a cumulative frequency distribution is called the cumulative frequency curve or ogive.

There are two methods to construct ogives :

(i) Less than ogive :

In this method, an ogive is cumulated upward. Scale the cumulative frequencies along the y-axis and exact upper limits along the x-axis. The scale along the y-axis should be such as may accommodate the total frequency.

Step I. Form the cumulative frequency table.

Step II. Mark the actual upper class limits along the x-axis.

Step III. Mark the cumulative frequency of the respective classes along the y-axis.

Step IV. Plot the points (upper limits, corresponding cumulative frequency).

By joining these points on the graph by a free hand curve, we get an ogive of 'less than' type.

(ii) More than ogive :

In this method, an ogive is cumulated downward. Scale the cumulative frequencies along the y-axis and the exact lower limits along the x-axis.

Step I. Scale the cumulative frequencies along the y-axis and the actual lower limits along the x-axis.

Step II. Plot the ordered pairs (lower limit, corresponding cumulative frequency). To complete an ogive, we also

plot the ordered pair (upper limit of the highest class,0).

Step III. Join these plotted points by a smooth curve. The curve so obtained is the required 'more than' type ogive.
Discussion
817.

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution 50.x: 1030507090y: 17f132f219

Answer»
Xyyx
1017170
30f130f1
50321600
70f270f2
90191710
 N = 120 \(\sum\)yx = 30f1 + 70f2 + 3480

Given, 

mean = 50

\(\frac{\sum yx}{N}=50\)

\(\frac{30f1+70f2+3480}{120}=50\)

30f1 + 70f2 + 3480 = 50 x 120

30f1 + 70f2 + 3480 = 6000 (i)

Also, ∑y = 120

17 + f1 + 32 + f2 + 19 = 120

f1 + f2 = 52

f1 = 52 – f2

Substituting value of f1 in (i), we get

30 (52 – f2) + 70f2 + 3480 = 6000

40f2 = 960

f2 = 24

Hence, 

f1 = 52 – 24 = 28

Therefore, 

f1 = 28 and f2 = 24
Discussion
818.

Find the difference between the upper limit of the median class and lower limit of the modal class.(you can take any example to teach)

Answer»

Let us consider the class 65-85.
Lower limt = 65
Uppr limit = 85

Difference between the upper limit and the lower limit = 85 - 65 = 20.
Hence, the answer is 20.
Discussion
819.

Will the median class and modal class of grouped data always be different? Justify your answer.

Answer»

The median class and modal class of grouped data is not always different, it depends on the data given.
Discussion
820.

Complete the following cumulative frequency table:Class(Height in cm)Frequency(No. of student)Less than type frequency150 - 153 0505153 -1560705 +....=.....156 - 15915... +15 =....159 - 16210....+....= 37162 - 1650537 + 5 = 42165 - 168 03....+..... = 45Total (N) = 45

Answer»

Required Table :

Class
(Height in cm)		Frequency
(No. of student)		Less than 
type frequency
150 - 153 0505
153 -1560705 + 07 = 12
156 - 1591512 + 15 = 27
159 - 1621027 + 10 = 37
162 - 1650537 + 5 = 42
165 - 168 0342 + 03 = 45
Total (N) = 45
Discussion
821.

Range of the series 25 , 33 , 44 , 26 , 17 is `"_______"`.

Answer» Correct Answer - 27
Discussion
822.

Range of 14, 12, 17, 18, 16 and x is 20. find x (x `gt` 0).A. 2B. 28C. 32D. Cannot be determined

Answer» Correct Answer - C
Range = Maximum value - Minimum value.
Discussion
823.

The weights of 20 students in a class are given below. Find the median of the above frequency distribution .A. 32.5B. 33C. 33.5D. 32

Answer» Correct Answer - b
Find the less than cumulative frequency , then find the mean by using formulae .
Discussion
824.

Calculate variance and standard deviation of the following data : 10 , 12 , 8 , 14 , 16 .

Answer» AM `(barx) = (10 + 12 + 8 + 14 + 16)/(5) = (60)/(5) = 12`
Varience = `((10 - 12)^(2) + (12-12)^(2) + (8-12)^(2) + (14-12)^(2) + (16-12)^(2))/(5)`
`= (4 + 0 + 16 + 4 + 16)/(5) = (40)/(5) = 8`
SD `(sigma) = sqrt("Variance") = sqrt(8)`.
Discussion
825.

What is the Range of the data x+1 , x+2 , x+6 ,2x+9 is 10 ,then find the value of x, `x gt 0` ?A. 2B. 8C. 7D. 0

Answer» Correct Answer - A
Discussion
826.

If a variable takes the discrete values `alpha-4`, `alpha -(7)/(2), alpha-(5)/(2), alpha-2,alpha+(1)/(2), alpha-(1)/(2), alpha+5(alpha gt 0)`, then the median isA. `alpha-(5)/(4)`B. `alpha-(1)/(2)`C. `alpha-2`D. `alpha+(5)/(4)`

Answer» Correct Answer - A
Arrange the data as follows :
`alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5`
Median `=(1)/(2)` [value of 4th item+value of 5th item]
`therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`
Discussion
827.

Find the coefficient of range of the scores 25, 30, 22, 34, 50, 56 and 67.

Answer» Correct Answer - `(45)/(89)`
Discussion
828.

Find the range of { 2 , 7 , 6 , 4 , 3 , 8 , 5 , 12}.

Answer» By arranging the given data in the ascending order .
We have , { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 12}
`therefore` Range = (Maximum value ) - (Minimumn value ) = 12 - 2= 10 .
Discussion
829.

Consider the data : 2 , x , 3 , 4 , 5 ,2 , 4, 6 ,4 , where x `gt` 2 . The mode of the data is `"_______"`.

Answer» Correct Answer - 4
Discussion
830.

Range of the scores 18 , 13 , 14 , 42 , 22 , 26 and x is 44 `(x gt 0)` . Find the sum of the digits of x .A. 16B. 14C. 12D. 18

Answer» Correct Answer - c
Range is the maximum value - minimum value .
Discussion
831.

Mode for the following distribution is 22 and 10 `gt y gt x` . Find y . (a) 2 (b) 5 (c) 3 (d) 4

Answer» (i) 10 is the model class .
(ii) Using mode = `L + (Delta_(1))/(Delta_(1) + Delta_(2)) xx C` , we can get the value of x .
Discussion
832.

The mean of the observations x – 1, x and x + 1 isA) 3x B) x C) 2x D) 0

Answer»

Correct option is (B) x

Mean \(=\frac{(x-1)+x+(x+1)}3\)

\(=\frac{3x}3=x\)

Correct option is  B) x
Discussion
833.

The difference between any two lower limits is called A) size of the class B) range C) frequencyD) upper limit

Answer»

Correct option is (A) size of the class

The difference between any two lower limits is called size of the class.

A) size of the class
Discussion