# Free Statistics 02 Practice Test - 10th Grade

In the formula ¯x=A+fidifi , the A stands for assumed mean.

A.

True

B.

False

#### SOLUTION

Solution : A

¯x=A+fidifi

In the formula, the A stands for the assumed mean.

The median class for the following data is
MarksBelow 10Below 20Below 30Below 40Below 50 No. of  students 510233140

A.

10 - 20

B.

20 - 30

C.

30 - 40

D.

40 - 50

#### SOLUTION

Solution : B

The given table can be written as:
Marks0101020203030404050 No. of students551389 Cumulative Frequency510233140
No. of students can be calculated as follows:
5
10 - 5 = 5
23 - 10 = 13
31 - 23 = 8
40 - 31 = 9

n2=402=20

Since there are a total of 40 observations, the median class is the one whose cumulative frequency is closest to and greater than 20.

Here the cumulative frequency for the class interval '10 - 20' is 23 which is greater than 20.

The class interval 20 - 30 is the required answer.

The median class for the given data is

Height (in cm) Less than 140 Less  than 145 Less than 150Less than 155Less than 160Less than 165No. of girls41129404651

A.

145 - 150

B.

150 - 155

C.

155 - 160

D.

160 - 165

#### SOLUTION

Solution : A

The given table can be written as:
Height (in cm) Less than 140 Less  than 145 Less than 150Less than 155Less than 160Less than 165No. of girls41129404651

Height (in cm) Less than 140140145145150150155155160160165Cumulative Frequency41129404651

n = 51

n2=512
= 25.5

Since there are a total of 51 observations, the median class is the one whose cumulative frequency is closest to and greater than 25.5. The cumulative frequency for the class '145 - 150' is 29 which is greater than 25.5. The median class is '140 - 145'. Thus, the class interval '145 - 150' is the required answer.

The median for the data is

IQ 60707080809090100100110110120120130No. of pupils2351614137

A.

102.86

B.

98.46

C.

97.47

D.

103.26

#### SOLUTION

Solution : A

ClassIntervalFrequencyCumulativeFrequency607022708035809051090100162610011014401101201353120130760

n2=30.
So, the median class is the one whose cumulative frequency is closest to and greater than 30.
the median class is 100-110

Median=l+(n2cff)×h
Where l is lower class limit of median class.
n is total number of observations.
cf is the cumulative frequency of the class preceding the median class.
f is the frequency of the median class and h is the class size.

Median=100+302614×10

Median=102.86

In the ___ method of calculating mean we divide the deviations by the same number to simplify calculations.

#### SOLUTION

Solution :

The method in question is the step-deviation method.

The average  marks  of  the students in Mathematics of class 8th is 45. But most of the students were given 60 out of 100.The Median of the class would be:

A.

30

B.

40

C.

50

D.

60

#### SOLUTION

Solution : C

Mean = 45 and Mode = 60
3 Median = Mode +2 Mean
3 Median = 60 +2(45)
3 Median = 150

Median=50

The mean ¯x of a grouped data is given by:

A.

fixifi

B.

fixi

C.

xifi

D.

fixix.f

#### SOLUTION

Solution : A

For grouped data, using the direct method first, we find the sum of the values of all the observations after multiplying them with their respective frequencies. Then we divide this result by the total number of observations (sum of all frequencies).

Which of the following represents an empirical relationship between Median, Mode and Mean?

A.

3 Mode= Median + 2 Mean

B.

Mode =3 Median - 2 Mean

C.

3 Mean= Median + 2 Mode

D.

can't say.

#### SOLUTION

Solution : B

The empirical relationship between Median, Mode and Mean is Mode = 3 Median - 2 Mean

Which of the following expressions is used to determine the mode?

A.

l+(f1f02f1f0f2) ×h

B.

1+(f1f0f1f0f2) ×h

C.

1+(f1f02f2f0f2) ×h

D.

2f1f02f1f0f2

#### SOLUTION

Solution : A

l+(f1f02f1f0f2) ×h is used to determine the mode .
A mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency.

A survey was conducted on 20 families in a locality by a group of students .What will be the mode of the data?

Age of family member02020404060608080100 Number of students78221

A.

21.85

B.

22.86

C.

23.87

D.

24.87

#### SOLUTION

Solution : B

Using, Mode =   l+f1f02f1f0f2×h
where,
Maximum class frequency, f1=8

The class corresponding to the frequency =2040

Lower limit of the modal class, l=20

Frequency of class preceding the modal class, f0=7

Frequency of class succeeding the modal class, f2=2

Now,Mode=20+871672×20

Mode=20+17×20

Mode=22.86