Statistics Test

Result

Statistics Test - 56

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The marks of the students of a class who appeared for a test in English are represented in the following frequency table:
Class
interval 1-10 11-20 21-30 31-40 41-50 51-60
Frequency 9 22 - 20 12 8 100
(Total frequency)
What is/are the modal class)es)?

A
$$10.5-20.5$$ only

B
$$20.5-30.5$$ only

C
$$10.5-20.5$$ and $$20.5-30.5$$

D
There is no modal class

Solution
Question 2
1 / -0

The median of three positive integers, no two of which are equal, is $$5$$. What is the least possible value of the arithmetic mean of these integers?

A
$$2$$

B
$$3$$

C
$$4$$

D
No such least possible value exists

Solution
Question 3
1 / -0

The mode of the following series is 36. Find the missing frequency in it
Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 10 ...... 16 12 6 7

A
$$10$$

B
$$15$$

C
$$16$$

D
$$12$$

Solution

We know that, Mode $$ = l+\frac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \times h $$
Where $$ l$$ is lower limit of the modal class, $$h$$ is size of the class interval, $${ f }_{ 1 } = $$ frequency of the modal class,$$ { f }_{ 0 } = $$frequency of the class preceding the modal class, $$ { f }_{ 2} = $$ frequency of the class succeeding the modal class
Since the class $$ 30-40 $$ has the highest frequency $$ 16 $$, it is the modal class.

Now, Mode $$ = 30 +\dfrac { 16-f }{ 32- f- 12 } \times 10 = 36 $$
$$ \implies 30 + \dfrac {16-f}{20-f}\times 10 = 36 $$
$$ \implies \dfrac {16-f}{20-f}\times 10 = 6 $$
$$ \implies 160 - 10f = 120 -6f $$
$$ \implies 40 = 4f $$
$$ \implies f = 10 $$
Question 4
1 / -0

Find the mode of the following data

A
$$16.5$$

B
$$17.5$$

C
$$18$$

D
$$18.5$$

Solution

CI Frequency(f)
0-10 5
10-20 8
20-30 7
30-40 6

40-50 4

As $$ 8 $$ is the highest frequency, modal class is $$ 10-20 $$ and the mode lies in this class

Now, Mode $$ = l+\dfrac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \times h $$

Where $${ l }_{ 1 } $$ lower limit of the modal class
$$ {f}_{1} = $$ frequency of the modal class
$$ {f}_{0} = $$ frequency of the class preceding the modal class
$$ {f}_{2} = $$ frequency of the class succeeding the modal class
$$ h = $$ size of the modal class

So, Mode $$ = 10+\dfrac { 8-5 }{ 2(8) - 5 - 7} \times 10 = 17.5 $$
Question 5
1 / -0

If the ratio of mean and median of a certain data is $$5:7$$, then find the ratio of its mode and mean.

A
$$2:5$$

B
$$11:5$$

C
$$6:5$$

D
$$2:3$$

Solution

Given: Mean $$:$$ Median $$= 5:7$$
To find: Mode $$:$$ Mean
For any given data, $$ \text{Mode = 3 Median - 2 Mean} $$
Let, Mean $$=5k$$ and median $$=7k$$
Now, mode $$=3\times 7k-2\times 5k=11k$$
$$\dfrac{\text{mode}}{\text{mean}}=\dfrac{11k}{5k}=\dfrac{11}{5}$$

$$\therefore$$ mode $$:$$ mean $$=11:5$$
Question 6
1 / -0

Directions For Questions

The average age of $$06$$ persons living in a house is $$23.5$$ years. Three of them are majors and their average age is $$42$$ years. The difference in ages of the three minor children is same.

...view full instructions

What is the median of the ages of minor children?

A
$$3$$ years

B
$$5$$ years

C
$$7$$ years

D
cannot determined due to the insufficient data

Solution
Question 7
1 / -0

For a grouped data, modal class is $$22-29.$$ If $$2f_0=f_1=3f_2$$ then Z=......,

A
$$23$$

B
$$24$$

C
$$25$$

D
$$26$$

Solution
Question 8
1 / -0

If a variable takes the discrete values $$\alpha + 4,\,\,\alpha - \frac{7}{2},\alpha - \frac{5}{2},\alpha - 3,\alpha - 2,\alpha + \frac{1}{2},\alpha - \frac{1}{2},\alpha + 5\left( {\alpha > 0} \right).$$ then the median is:

A
$$\alpha - \frac{5}{4}$$

B
$$\alpha - \frac{1}{2}$$

C
$$\alpha - 2$$

D
$$\alpha + \frac{5}{4}$$

Solution
Question 9
1 / -0

.

A
5

B
5.28

C
6

D
6.29

Solution
Question 10
1 / -0

The median and S.D of a distribution are 20 and 4 respectively. If each item is increased by 2, the new median are S.D are:

A
20, 6

B
22. 6

C
18, 6

D
22, 4

Solution