# Free Statistics 03 Practice Test - 10th Grade

### Question 1

The median for grouped data is formed by using the formula:

l+(n2−cff ) ×h

l−(n2−cff ) ×h.

l+(n2+cff ) ×h.

l+2n

#### SOLUTION

Solution :A

The median for grouped data is formed by 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.

### Question 2

The following data shows monthly savings of 100 families . Calculate the mode of the given frequency distribution.

Monthly savings(Rs) Number of families 1000−2000142000−3000153000−4000214000−5000275000−600025

4790

4760

4750

4780

#### SOLUTION

Solution :C

Modal class is 4000-5000 within

f1 = 27, f0 = 21, f2 = 25, I = 4000, h = 1000

Mode = l + f1−f02f1−f0−f2×h

Mode = 4000 + 68×1000

= 4750

### Question 3

Find the mode of the following data.

Class interval0−100100−200200−300300−400400−500Frequency7111598

#### SOLUTION

Solution :Modal Class is 200-300

Mode=I+(f1−f02f1−f0−f2)×h

=200+15−1130−11−9×100=240

### Question 4

The below table shows that profit made by a group of shops in mall. Then the median is _____.

Profit per shop less than (%)102030405060No. of shops1230577794100

27.4

31.2

26.8

23.7

#### SOLUTION

Solution :A

Firstly, the cumulative frequency (CF) table is drawn.

C.ICFFrequency0−10121210−20301820−30572730−40772040−50941750−601006Total100

n2=1002=50

C.F nearest to 50 and greater than 50 is 57.

∴ 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=20+(50−3027)×10Median = 27.4

### Question 5

The times (in seconds) taken by 150 athelets to run a 110 m hurdle race are tabulated below

Class13.8−1414−14.214.2−14.414.4−14.614.6−14.814.8−15Frequency245714820

The number of atheletes who completed the race in less than 14.6 s is

11

71

82

130

#### SOLUTION

Solution :C

The number of atheletes who completed the race in less than 14.6

= 2 + 4 + 5 + 71

= 82

### Question 6

In the assumed mean method, if A is the assumed mean, then deviation di is :

#### SOLUTION

Solution :A

The deviation is di=xi−A

### Question 7

In the assumed mean method, if A is the assumed mean, then deviation di is :

xi−A

xi+A

xi

A−xi

#### SOLUTION

Solution :A

In statistics, 'the assumed mean' is a method for calculating the arithmetic mean and standard deviation of a data set. It simplifies calculating accurate values by hand.

During the application of the short-cut method for finding the mean, the deviation d, are divisible by a common number ‘h’. In this case, the di = xi – A is reduced to a great extent as 'di' becomes dih. In this method, we divide the deviations by the same number to simplify calculations. The deviation is di=xi−A

### Question 8

For which data set, the mean is not a good representative measures of central tendency.

{10, 12, 15, 17, 14, 15, 17}

{2, 20, 25, 17, 19, 22}

{5, 7, 5, 6, 4, 5, 7}

{17, 20, 25, 17, 19, 22}

#### SOLUTION

Solution :B

We know that extreme values in the data affect the mean. Here only for the data set {2, 20, 25, 17, 19, 22}, we find the extreme value as one class has the frequency as 2 and the others have frequency 20, 25, 17, 19 and 22.

### Question 9

The following table gives the life time of 200 neon lamps. Find the median class.

#### SOLUTION

Solution :A

The median class is identified as the class whose cumulative frequency will be greater than n2, where 'n' is the sum of the frequencies. (n=200).

Next, we divide 200 by 2.

n2=2002=100

We now locate the class whose cumulative frequency is greater than 100.

"2500 - 3000" is the class whose cumulative frequency 155 is greater than 100.

∴The median class is 2500 - 3000.

### Question 10

#### SOLUTION

Solution :B

Here we locate a class with the maximum frequency. From the given table we find the maximum frequency as 12 whose class is 2 - 3. Therefore the modal class is 2 - 3.