# Free Statistics 01 Practice Test - 10th Grade

Class mark of a class is _____

A.

Upper limit + lower limit

B.

upper limit + lower limit2

C.

upper limit - lower limit2

D.

upper limit - lower limit

#### SOLUTION

Solution : B

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

The mean of the following distribution is ____.
xi10131619fi2576

A.

15.2

B.

15.35

C.

15.55

D.

16

#### SOLUTION

Solution : C

xififixi1022013565167112196114fi=20 fixi=311

Mean =fixifi

=31120

=15.55

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

___

xi10p182125fi1015799

#### SOLUTION

Solution :

xififixi1010100p1515p187126219189259225

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

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

A. 6
B.

7

C.

8

D.

9

#### SOLUTION

Solution : 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

Mode is

A.

Most frequent value

B.

Least frequent value

C.

Middle most value

D.

Average value

#### SOLUTION

Solution : A

Mode is the most frequently observed value

In the fromula for mode of a grouped data ,

mode =l+[t1t02f1f0f2]×h ,

where symbols have their usual meaning f1   represents

A.

Frequency of Modal class

B.

Frequency of the class preceding the modal class

C.

Frequency of the class succeeding the modal class

D.

Frequency of median class

#### SOLUTION

Solution : A

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

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

A.

True

B.

False

#### SOLUTION

Solution : 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

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.

A.

True

B.

False

#### SOLUTION

Solution : 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.

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

A.

12

B.

14

C.

15

D.

18

#### SOLUTION

Solution : B

We know Mean = fxf

Hence, 5x+212 = 6

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

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

A. 4
B. 3
C. 6
D. 7

#### SOLUTION

Solution : A

We know that,

3 Median = Mode + 2 Mean
or
Mode = 3 Median - 2 Mean
=  3×62×7
= 18 - 14
= 4

Therefore, mode = 4