# Free Statistics 02 Practice Test - 10th Grade

### Question 1

In the formula ¯x=A+∑fidi∑fi , the A stands for assumed mean.

True

False

#### SOLUTION

Solution :A

¯x=A+∑fidi∑fi

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

### Question 2

The median class for the following data is

MarksBelow 10Below 20Below 30Below 40Below 50 No. of students 510233140

10 - 20

20 - 30

30 - 40

40 - 50

#### SOLUTION

Solution :B

The given table can be written as:

Marks0−1010−2020−3030−4040−50 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.

### Question 3

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

145 - 150

150 - 155

155 - 160

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 140140−145145−150150−155155−160160−165Cumulative 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.

### Question 4

The median for the data is

IQ 60−7070−8080−9090−100100−110110−120120−130No. of pupils2351614137

102.86

98.46

97.47

103.26

#### SOLUTION

Solution :A

ClassIntervalFrequencyCumulativeFrequency60−702270−803580−9051090−1001626100−1101440110−1201353120−130760

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+(n2−cff)×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+30−2614×10

∴ Median=102.86

### Question 5

In the

#### SOLUTION

Solution :The method in question is the step-deviation method.

### Question 6

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:

30

40

50

60

#### SOLUTION

Solution :C

Mean = 45 and Mode = 60

3 Median = Mode +2 Mean

3 Median = 60 +2(45)

3 Median = 150

Median=50

### Question 7

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

∑fixi∑fi

∑fixi

∑xifi

∑fixi∑x.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).

### Question 8

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

3 Mode= Median + 2 Mean

Mode =3 Median - 2 Mean

3 Mean= Median + 2 Mode

can't say.

#### SOLUTION

Solution :B

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

### Question 9

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

l+(f1−f02f1−f0−f2) ×h

1+(f1−f0f1−f0−f2) ×h

1+(f1−f02f2−f0−f2) ×h

2f1−f02f1−f0−f2

#### SOLUTION

Solution :A

l+(f1−f02f1−f0−f2) ×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.

### Question 10

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 member0−2020−4040−6060−8080−100 Number of students78221

21.85

22.86

23.87

24.87

#### SOLUTION

Solution :B

Using, Mode = l+f1−f02f1−f0−f2×h

where,

Maximum class frequency, f1=8The class corresponding to the frequency =20−40

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+8−716−7−2×20

Mode=20+17×20

∴Mode=22.86